General Physics II ELECTRICITY AND MAGNETISM

General Physics II ELECTRICITY AND MAGNETISM
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General Physics II: Electricity & Magnetism
Course Description Coulomb's law, the electrostatic field, Gauss’s Law, the electrostatic potential, capacitance and dielectrics, electric current, resistance and electromotive force, direct current circuits, magnetic field and magnetic forces, sources of magnetic fields, Ampere's Law, Faraday's Law, induction and Maxwell's equations. II. Course Objectives To provide a foundation in physics necessary for further study in science, engineering and technology. To provide an appreciation of the nature of physics, its methods and its goals. To contribute to the development of the student's thinking process through the understanding of the theory and application of this knowledge to the solution of practical problems.

Textbook: the class notes beside the following textbooks:
Physics for Scientists and Engineers, Raymond A. Serway, 6th Edition University Physics, Sears, Zemansky and Young Course Outline Charge and Matter: Charge and conservation of charges, Material and charge, electric forces, Coulombs Law The Electric Field: Electric field, The lines of forces, Electric dipole, Continuous charge distribution, Effect of electric field on point charge, Milliken’s Experiment. Electric Flux and Gauss’s Law: Flux of electric field, Gauss’s Law and its application.

Electric Potential: Definition of electric potential, Potential difference between two points, Calculation of electric potential, Electric potential energy, Electric field and potential. Capacitors: Capacitor and capacitance, Parallel plate capacitor, Cylindrical capacitor, Spherical capacitor, Capacitors connections, Energy stored in capacitor, Effect of insulator inside a capacitor. Electric current and Ohm’s Law: Electric current, Current density, Resistivity, Ohm’s Law, Electric power, Electromotive force and electric circuits, Kirchoff´s Law, RC circuit The Magnetic Field: Definition and introduction, Flux of magnetic field, Magnetic force, Hall effect, Torque on a current loop, charge in a magnetic field, Biot-Savart Law, Helmholtz coils, Ampere’s Law.

Faraday's Law : Faraday's Law of Induction, Motional emf, Lenz's Law, Induced emf and Electric Fields, Generators and Motors/ Eddy Currents, Maxwell's Equations Inductance : Self-Inductance, RL Circuits, Energy in a Magnetic Field, Mutual Inductance, Oscillations in an LC Circuit, The RLC Circuit. Alternating Current Circuits :AC Sources, Resistors in an AC Circuit, Inductors in an AC Circuit, Capacitors in an AC Circuit, The RLC Series Circuit. Power in an AC Circuit, Resonance in a Series RLC Circuit, The Transformer and Power Transmission, Rectifiers and Filters.

GRADING POLICY Weight Grade Component 20 HW/Quizzes Midterm Exam 60
Your grade will be judged on your performance in Home work, Quizzes, tow tests and the Lab. Points will be allocated to each of these in the following manner: GRADING SCALE: Weight Grade Component 20 HW/Quizzes Midterm Exam 60 Final Exam 100 Total

Lecture I Electrostatic

Introduction Knowledge of electricity dates back to Greek antiquity (700 BC). Began with the realization that amber when rubbed with wool, attracts small objects. This phenomenon is not restricted to amber/wool but may occur whenever two non-conducting substances are rubbed together.

Net Electrical Charge Matters are made of atoms. An atom is basically composed of three different components : electrons, protons, and neutrons. An electron can be removed easily from an atom Normally, an atom is electrically neutral, which means that there are equal numbers of protons and electrons. Positive charge of protons is balanced by negative charge of electrons. It has no net electrical charge. When atoms gain or lose electrons, they are called "ions." A positive ion is a cation that misses electrons. A negative ion is an anion that gains extra electrons.

What is charge? Objects that exert electric forces are said to have charge. Charge is the source of electrical force. There are two kinds of electrical charges, positive and negative. Same charges (+ and +, or - and -) repel and opposite charges (+ and -) attract each other.

Charged Objects The Law of Conservation of Charge
The Law of conservation of charge states that the net charge of an isolated system remains constant. Charged Objects When two objects are rubbed together, some electrons from one object move to another object. For example, when a plastic bar is rubbed with fur, electrons will move from the fur to the plastic stick. Therefore, plastic bar will be negatively charged and the fur will be positively charged.

"charge is quantized" in terms of an equation, we say:
Quantization Robert Millikan found, in 1909, that charged objects may only have an integer multiple of a fundamental unit of charge. Charge is quantized. An object may have a charge ±e, or ± 2e, or ± 3e, etc but not say ± 1.5e. Proton has a charge +1e. Electron has a charge –1e. Some particles such a neutron have no (zero) charge. "charge is quantized" in terms of an equation, we say: q = n e

Unit of Electrical Charge:
The Coulomb " C " The symbol for electric charge is written q, - q or Q. The unit of electric charge is coulomb "C". The charge of one electron is equal to the charge of one proton, which is 1.6 * C. This number is given a symbol "e". Example: How many electrons are there in 1 C of charge?

Insulators and Conductors( Material classification)
Materials/substances may be classified according to their capacity to carry or conduct electric charge Conductors are material in which electric charges move freely. Insulator are materials in which electrical charge do not move freely. Glass, Rubber are good insulators. Copper, aluminum, and silver are good conductors. Semiconductors are a third class of materials with electrical properties somewhere between those of insulators and conductors. Silicon and germanium are semiconductors used widely in the fabrication of electronic devices.

Why? is static electricity more apparent in winter?
Example: Identify substances or materials that can be classified as Conductors ? Insulators? Why? is static electricity more apparent in winter?

Lecture 2 Coulomb’s law

Coulomb’s Law Coulomb discovered in 1785 the fundamental law
of electrical force between two stationary charged particles. An electric force has the following properties: Inversely proportional to the square of the separation, r, between the particles, and is along a line joining them. Proportional to the product of the magnitudes of the charges |q1| and |q2| on the two particles. Attractive if the charges are of opposite sign and repulsive if the charges have the same sign.

Coulomb’s Law (Mathematical Formulation)
ke known as the Coulomb constant. Value of ke depends on the choice of units. SI units Force: the Newton (N) Distance: the meter (m). Charge: the coulomb ( C).

Experimentally measurement: ke = 8.9875×109 Nm2/C2.
where eo is known as the Permittivity constant of free space. eo = 8.85 x C2/N.m2 Experimentally measurement: ke = ×109 Nm2/C2. Reasonable approximate value: ke = 8.99×109 Nm2/C2.

Example: the Coulomb constant unit
Then the Coulomb constant unit is

The electrostatic force
The electrostatic force is often called Coulomb force. It is a force (thus, a vector): a magnitude a direction. + - r q1 q2 F21 F12 + r q1 q2 F21 F12

Example: The electron and proton of a hydrogen atom are separated (on the average) by a distance of about 5.3x10-11 m. Find the magnitude of the electric force that each particle exerts on the other. q1 =-1.60x10-19 C q2 =1.60x10-19 C r = 5.3x10-11 m Attractive force with a magnitude of 8.2x10-8 N.

® ® r Gravitational vs. Electrical Force For an electron: F = 1 4 pe q
m1 q2 m2 F elec = 1 4 pe q 2 r grav = G m F elec grav = q 1 2 m 4 pe G For an electron: |q| = 1.6 × C m = 9.1 × kg F elec grav = × + 4 17 10 42 .

Superposition of Forces
+Q1 +Q2 +Q0 +Q3

The net force on q3 is the vector sum of the
forces F32 and F31. The magnitude of the forces F32 and F31 can calculated using Coulomb’s law.

Example: Two fixed charge, 1µC and -3µC are separated by 10 cm
Example: Two fixed charge, 1µC and -3µC are separated by 10 cm. Where may a third charge be located so that no force act on it ? Solve the eq. to find d.

Examination of the geometry of Figure leads to
If L is much larger than x (which is the case if Ө is very small), we may neglect x/2 in the denominator and write tan Ө = x/2L. This is equivalent to approximating tanӨ by sinӨ. The magnitude of the electrical force of one ball on the other is by Eq. When these two expressions are used in the equation mg tan Ө = Fe, we obtain

b) We solve x3 for the charge:
Thus, the magnitude is

Lecture 3 The Electric Field

Electric Field Suggests the notion of electrical field (first introduced by Michael Faraday ( ). An electric field is said to exist in a region of space surrounding a charged object. If another charged object enters a region where an electrical field is present, it will be subject to an electrical force.

Electric Field & Electric Force
Consider a small charge q0 near a larger charge Q. We define the electric field E at the location of the small test charge as a ratio of the electric force F acting on it and the test charge q0 This is the field produced by the charge Q, not by the charge q0

Electric Field Direction
The direction of E at a point is the direction of the electric force that would be exerted on a small positive test charge placed at that point. E E - - - + +

Electric Field from a Point Charge
Suppose we have two charges, q and q0, separated by a distance r. The electric force between the two charges is We can consider q0 to be a test charge, and determine the electric field from charge q as

If q is +ve, field at a given point is radially outward
from q. r E + qo q If q is -ve, field at a given point is radially inward from q. r - E qo q

Electric Field Lines To visualize electric field patterns, one can draw lines pointing in the direction of the electric field vector at any point. These lines are called electric field lines. The electric field vector is tangent to the electric field lines at each point. The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region. No two field lines can cross each other . Why?

The electric field lines for a point charge.
(a) For a positive point charge, the lines are directed radially outward. (b) For a negative point charge, the lines are directed radially inward. Note that the figures show only those field lines that lie in the plane of the page.

The electric field lines for two positive point charges.
The electric field lines for two point charges of equal magnitude and opposite sign (an electric dipole)

Question: Two charges q1 and q2, fixed along the x-axis as shown, produce an electric field E at the point (x,y)=(0,d), which is the directed along the negative y-axis. Which of the following is true? Both charges are positive Both charges are negative The charges have opposite signs

Electric Field from an Electric Dipole
A system of two oppositely charged point particles is called an electric dipole. The vector sum of the electric field from the two charges gives the electric field of the dipole (superposition principle). We have shown the electric field lines from a dipole

Example: Two charges on the x-axis a distance d apart Put -q at x = -d/2 Put +q at x = +d/2 Calculate the electric field at a point P a distance x from the origin

Replacing r+ and r- we get
Principle of superposition:The electric field at any point x is the sum of the electric fields from +q and -q Replacing r+ and r- we get This equation gives the electric field everywhere on the x-axis (except for x = d/2)

Example: Electric Field Due to Two Point Charges
Charge q1=7.00 mC is at the origin, and charge q2= mC is on the x axis, m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. x y 0.300 m q1 q2 0.400 m P E1 E2 E

In Figure, determine the point (other than infinity) at which the
Example In Figure, determine the point (other than infinity) at which the total electric field is zero. Solution: The sum of two vectors can be zero only if the two vectors have the same magnitude and opposite directions.

Motion of charge particles in a uniform electric field
An electron moving horizontally passes between two horizontal planes, the upper plane charged negatively, and the lower positively. A uniform, upward-directed electric field exists in this region. This field exerts a force on the electron. Describe the motion of the electron in this region. ve -

Horizontally: Vertically: No electric field No force No acceleration
Constant horizontal velocity Vertically: Constant electric field Constant force Constant acceleration Vertical velocity increase linearly with time.

The charge will follow a parabolic path downward.
- Conclusions: The charge will follow a parabolic path downward. Motion similar to motion under gravitational field only except the downward acceleration is now larger.

This device is known as a cathode ray tube (CRT)
-Q +Q Phosphor Screen This device is known as a cathode ray tube (CRT)

Continuous Charge Distributions
Single charge Single piece of a charge distribution + +Q3 +Q2 +Q1 Discrete charges Continuous charge distribution

Cartesian Polar Line charge Surface charge Volume charge

Example: Electric Field Due to a Charged Rod
A rod of length l has a uniform positive charge per unit length λ and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end.

Example – Infinitely Long Line of Charge
+ + y-components cancel by symmetry + + + + + +

Example – Charged Ring perpendicular-components cancel by symmetry + +

When: The charged ring must look like a point source.

Example – Uniformly Charged Disk

Two Important Limiting Cases
Large Charged Plate: Very Far From the Charged Plate:

Lecture 4 Discussion

[1] In figure, two equal positive charges q=2x10-6C interact with a third charge Q=4x10-6C. Find the magnitude and direction of the resultant force on Q.

[2] A charge Q is fixed at each of two opposite corners of a square as shown in figure. A charge q is placed at each of the other two corners. If the resultant electrical force on Q is Zero, how are Q and q related.

[3] Two fixed charges, 1µC and -3µC are separated by 10cm as shown in figure (a) where may a third charge be located so that no force acts on it? (b) is the equilibrium stable or unstable for the third charge?

[4] Find the electric field at point p in figure due to the charges shown.
Solution:

[5] A charged cord ball of mass 1g is suspended on a light string in the presence of a uniform electric field as in figure. When E=(3i+5j) *105N/C, the ball is in equilibrium at Θ=37o. Find (a) the charge on the ball and (b) the tension in the string. Substitute T from equation (1) into equation (2) Substitute T from equation (1) into equation (2) Substitute by q into equation (1) to find T=5.44*10-3N

[6] A 1. 3µC charge is located on the x-axis at x=-0. 5m, 3
[6] A 1.3µC charge is located on the x-axis at x=-0.5m, 3.2µC charge is located on the x-axis at x=1.5m, and 2.5µC charge is located at the origin. Find the net force on the 2.5µC charge

[7] Two free point charges +q and +4q are a distance 1cm apart
[7] Two free point charges +q and +4q are a distance 1cm apart. A third charge is so placed that the entire system is in equilibrium. Find the location, magnitude and sign of the third charge. Is the equilibrium stable?

[8] Two protons in a molecule are separated by a distance of 3. 8
[8] Two protons in a molecule are separated by a distance of 3.8*10-10m. Find the electrostatic force exerted by one proton on the other. [9] The electric force on a point charge of 4.0mC at some point is 6.9*10-4N in the positive x direction. What is the value of the electric field at that point?

[10] Two point charges are a distance d apart
[10] Two point charges are a distance d apart . Find E points to the left P. Assume q1=+1.0*10-6C, q2=+3.0*10-6C, and d=10cm

[11] Calculate E (direction and magnitude) at point P in Figure.

[12] A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 2.0cm away, in a time 1.5*10-8s. (a) What is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field .

[13] Three charges are placed on corners of an equilateral triangle as shown in Figure 1. An electron is placed at the center of the triangle. What is the magnitude of the net force on the electron?

[14] A uniform electric field exists in the region between two oppositely charged plane parallel plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate 2x10-8 s later. If the magnitude of the electric field is 4x103 N/C, what is the separation between the plates?

Lecture 5 Electric Flux and Gauss’s Law

Electric Flux Electric flux quantifies the notion “number of field lines crossing a surface.” The electric flux  through a flat surface in a uniform electric field depends on the field strength E, the surface area A, and the angle  between the field and the normal to the surface. Mathematically, the flux is given by Here A is a vector whose magnitude is the surface area A and whose orientation is normal to the surface. E A q

When q < 90˚, the flux is positive (out of the surface), and when q > 90˚, the flux is negative. Units: Nm2/C in SI units, the electric flux is a SCALAR quantity Find the electric flux through the area A = 2 m2, which is perpendicular to an electric field E=22 N/C Answer: F = 44 Nm2/C.

Example: Calculate the flux of a constant E field (along x) through a cube of side “L”. y 2 1 Solution E x z

Question: The flux through side B of the cube in the figure is the
same as the flux through side C. What is a correct expression for the flux through each of these sides? Click to add notes

When we have a complicated surface, we can divide it up into tiny elemental areas:

Example: What’s the total flux on a closed surface with a charge inside? The shape and size don’t matter! Just use a sphere

What is Gauss’s Law? Gauss’s Law does not tell us anything new, it is NOT a new law of physics, but another way of expressing Coulomb’s Law Gauss’s law makes it possible to find the electric field easily in highly symmetric situations.

Gauss’ Law The precise relation between flux and the enclosed charge is given by Gauss’ Law e0 is the permittivity of free space in the Coulomb’s law The symbol has a little circle to indicate that the integral is over a closed surface. The flux through a closed surface is equal to the total charge contained divided by permittivity of free space

A few important points on Gauss’ Law
The integral is over the value of E on a closed surface of our choice in any given situation The charge Qencl is the net charge enclosed by the arbitrary close surface of our choice. It does NOT matter where or how much charge is distributed inside the surface The charge outside the surface does not contribute. Why?

Question What’s the total flux with the charge outside? Why? Solution:
Zero. Because the surface surrounds no charge

Gauss Coulomb Calculate E of point like (+) charge Q
Consider sphere radius r centered at the charge Spherical symmetry: E is the same everywhere on the sphere, perpendicular to the sphere

Application of Gauss’s law Gauss’s law can be used to calculate the electric field if the symmetry of the charge distribution is high. Here we concentrate in three different ways of charge distribution A linear charge distribution A surface charge distribution A volume charge distribution

A linear charge distribution
Let’s calculate the electric field from a conducting wire with charge per unit length  using Gauss’ Law We start by assuming a Gaussian surface in the form of a right cylinder with radius r and length L placed around the wire such that the wire is along the axis of the cylinder

From symmetry we can see that the electric field will extend radially from the wire.
How? If we rotate the wire along its axis, the electric field must look the same Cylindrical symmetry If we imagine a very long wire, the electric field cannot be different anywhere along the length of the wire Translational symmetry Thus our assumption of a right cylinder as a Gaussian surface is perfectly suited for the calculation of the electric field using Gauss’ Law.

The electric flux through the ends of the cylinder
is zero because the electric field is always parallel to the ends. The electric field is always perpendicular to the wall of the cylinder so … and now solve for the electric field

Assume that we have a thin, infinite
A surface charge distribution Assume that we have a thin, infinite non-conducting sheet of positive charge The charge density in this case is the charge per unit area,  From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet

To calculate the electric field using Gauss’ Law,
we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height 2r, chosen to cut through the plane perpendicularly. Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the ends of the cylinder. Using Gauss’ Law we get … so the electric field from an infinite non-conducting sheet with charge density 

Assume that we have a thin, infinite conductor
(metal plate) with positive charge The “charge density” in this case is also the charge per unit area, , on either surface; there is equal surface charge on both sides. From symmetry, we can see that the electric field will be perpendicular to the surface of the sheet To calculate the electric field using Gauss’ Law, we assume a Gaussian surface in the form of a right cylinder with cross sectional area A and height r, chosen to cut through one side of the plane perpendicularly.

The field inside the conductor is zero so the end inside the
conductor does not contribute to the integral. Because the electric field is perpendicular to the plane everywhere, the electric field will be parallel to the walls of the cylinder and perpendicular to the end of the cylinder outside the conductor. Using Gauss’ Law we get … so the electric field from an infinite conducting sheet with surface charge density  is NB: A tricky factor of 2 sigma_{surface} = ½ sigma_{2 surfaces} So, E = sigma_{combined} / (2 eps_0)

let’s calculate the electric field from
A volume charge distribution let’s calculate the electric field from charge distributed uniformly throughout charged sphere. Assume that we have insolating a solid sphere of charge Q with radius r with constant charge density per unit volume . We will assume two different spherical Gaussian surfaces r2 > r (outside) r1 < r (inside)

Let’s start with a Gaussian surface with r1 < r.
From spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. Gauss’ Law gives us Solving for E we find inside

In terms of the total charge Q …
inside

Now consider a Gaussian surface with radius r2 > r.
Again by spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. Gauss’ Law gives us Solving for E we find outside same as a point charge!

Electric field vs. radius for a conducting sphere

Properties of Conductors
E is zero within conductor If there is a field in the conductor, then the free electrons would feel a force and be accelerated. They would then move and since there are charges moving the conductor would not be in electrostatic equilibrium. Thus E=0 net charge within the surface is zero How ?

Lecture 6 Application (Gauss’s Law)

[1] A solid conducting sphere of radius a has a net charge +2Q
[1] A solid conducting sphere of radius a has a net charge +2Q. A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge –Q as shown in figure. Using Gauss’s law find the electric field in the regions labeled 1, 2, 3, 4 and find the charge distribution on the spherical shell. Region (1) r < a To find the E inside the solid sphere of radius a we construct a Gaussian surface of radius r < a E = 0 since no charge inside the Gaussian surface Region (3) b > r < c E=0 How?

Region (2) a < r < b we construct a spherical Gaussian surface of radius r Region (4) r > c we construct a spherical Gaussian surface of radius r > c, the total net charge inside the Gaussian surface is q = 2Q + (-Q) = +Q Therefore Gauss’s law gives

[2] A long straight wire is surrounded by a hollow cylinder whose axis coincides with that wire as shown in figure. The solid wire has a charge per unit length of +λ , and the hollow cylinder has a net charge per unit length of +2λ . Use Gauss law to find (a) the charge per unit length on the inner and outer surfaces of the hollow cylinder and (b) the electric field outside the hollow cylinder, a distance r from the axis. (a) Use a cylindrical Gaussian surface S1 within the conducting cylinder where E=0

(b) For a Gaussian surface S2 outside the conducting cylinder

[3] Consider a long cylindrical charge distribution of radius R with a uniform charge density ρ. Find the electric field at distance r from the axis where r < R. If we choose a cylindrical Gaussian surface of length L and radius r, Its volume is πr²L , and it encloses a charge ρπr²L . By applying Gauss’s law we get, radially outward from the cylinder axis Thus Notice that the electric field will increase as r increases, and also the electric field is proportional to r for r<R. For the region outside the cylinder (r>R), the electric field will decrease as r increases.

Two Parallel Conducting Plates
When we have the situation shown in the left two panels (a positively charged plate and another negatively charged plate with the same magnitude of charge), both in isolation, they each have equal amounts of charge (surface charge density s) on both faces. But when we bring them close together, the charges on the far sides move to the near sides, so on that inner surface the charge density is now 2s. A Gaussian surface shows that the net charge is zero (no flux through sides — dA perpendicular to E, or ends — E = 0). E = 0 outside, too, due to shielding, in just the same way we saw for the sphere.

Two Parallel Nonconducting Sheets
The situation is different if you bring two nonconducting sheets of charge close to each other. In this case, the charges cannot move, so there is no shielding, but now we can use the principle of superposition. In this case, the electric field on the left due to the positively charged sheet is canceled by the electric field on the left of the negatively charged sheet, so the field there is zero. Likewise, the electric field on the right due to the negatively charged sheet is canceled by the electric field on the right of the positively charged sheet. The result is much the same as before, with the electric field in between being twice what it was previously.

We know previously that for each sheet,
[4] Two large non-conducting sheets of +ve charge face each other as shown in figure. What is E at points (i) to the left of the sheets (ii) between them and (iii) to the right of the sheets? We know previously that for each sheet, the magnitude of the field at any point is a) At point to the left of the two parallel sheets

b) At point between the two sheets
(c) At point to the right of the two parallel sheets

[5] A square plate of copper of sides 50cm is placed in an extended electric field of 8*104N/C directed perpendicular to the plate. Find (a) the charge density of each face of the plate

[6] An electric field of intensity 3. 5. 103N/C is applied the x axis
[6] An electric field of intensity 3.5*103N/C is applied the x axis. Calculate the electric flux through a rectangular plane 0.35m wide and 0.70m long if (a) the plane is parallel to the yz plane, (b) the plane is parallel to the xy plane, and (c) the plane contains the y axis and its normal makes an angle of 40o with the x axis. (a) the plane is parallel to the yz plane (b) the plane is parallel to the xy plane The angel 90 c) the plane is parallel to the xy plane The angel 40

[7] A long, straight metal rod has a radius of 5cm and a charge per unit length of 30nC/m. Find the electric field at the following distances from the axis of the rod: (a) 3cm, (b) 10cm, (c) 100cm.

[8] The electric field everywhere on the surface of a conducting hollow sphere of radius 0.75m is measured to be equal to 8.90*102N/C and points radially toward the center of the sphere. What is the net charge within the surface?

[9] A point charge of +5mC is located at the center of a sphere with a radius of 12cm. What is the electric flux through the surface of this sphere? [10] (a) Two charges of 8mC and -5mC are inside a cube of sides 0.45m. What is the total electric flux through the cube? (b) Repeat (a) if the same two charges are inside a spherical shell of radius m.

[12] A solid copper sphere 15cm in radius has a total charge of 40nC
[12] A solid copper sphere 15cm in radius has a total charge of 40nC. Find the electric field at the following distances measured from the center of the sphere: (a) 12cm, (b) 17cm, (c) 75cm. (a) At 12 cm the charge in side the Gaussian surface is zero so the electric field E=0 (b)

[13] Two long, straight wires are separated by a distance d = 16 cm, as shown below. The top wire carries linear charge density 3 nC/m while the bottom wire carries -5 nC/m. 1- What is the electric field (including direction) due to the top wire at a point exactly half-way between the two wires? 2- Find the electric field due to the bottom wire at the same point, exactly half-way between the two wires (including direction). 3- Work out the total electric field at that point.

using Gauss’ Law and cylindrical Gaussian surface as Lec.5 page 15

The Electric Potential
Lecture 7 The Electric Potential

Electric Potential Energy
The electric force, like the gravitational force, is a conservative force. (Conservative force: The work is path-independent.) As in mechanics, work is Work done on the positive charge by moving it from A to B A B d

The work done by a conservative force equals
the negative of the change in potential energy, DPE This equation is valid only for the case of a uniform electric field If a charged particle moves perpendicular to electric field lines, no work is done. if d  E

Electric potential is a scalar quantity
The potential difference between points A and B, VB-VA, is defined as the change in potential energy (final minus initial value) of a charge, q, moved from A to B, divided by the charge Electric potential is a scalar quantity Electric potential difference is a measure of electric energy per unit charge Potential is often referred to as “voltage” If the work done by the electric field is zero, then the electric potential must be constant

Electric potential difference is the work done to
move a charge from a point A to a point B divided by the magnitude of the charge. Thus the SI units of electric potential difference In other words, 1 J of work is required to move a 1 C of charge between two points that are at potential difference of 1 V Question: How can a bird stand on a high voltage line without getting zapped?

Units of electric field (N/C) can be expressed in
terms of the units of potential (as volts per meter) Because the positive tends to move in the direction of the electric field, work must be done on the charge to move it in the direction, opposite the field. Thus, A positive charge gains electric potential energy when it is moved in a direction opposite the electric field A negative charge looses electrical potential energy when it moves in the direction opposite the electric field

Example : A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A +12µC charge moves from the origin to the point (x,y) = (20cm, 50cm). (a) What was the change in the potential energy of this charge? (b) Through what potential difference did the charge move? Begin by drawing a picture of the situation, including the direction of the electric field, and the start and end point of the motion. (a) The change potential energy is given by the charge times the field times the distance moved parallel to the field. Although the charge moves 50cm in the y direction, the y direction is perpendicular to the field. Only the 20cm moved parallel to the field in the x direction matters for determining the change of potential energy. ∆ PE = -qEd = -(+12µC)(250 V/m)(0.20 m) = -6.0×10-4 . (b) The potential difference is the difference of electric potential,. ∆ V = ∆ PE / q = -6.0×10-4 J / 12µC = -50 V.

Analogy between electric and gravitational fields
The same kinetic-potential energy theorem works here If a positive charge is released from A, it accelerates in the direction of electric field, i.e. gains kinetic energy If a negative charge is released from A, it accelerates in the direction opposite the electric field A A d d q m B B

Example: motion of an electron
What is the speed of an electron accelerated from rest across a potential difference of 100V? Vab Given: DV=100 V me = 9.11×10-31 kg mp = 1.67×10-27 kg |e| = 1.60×10-19 C Find: ve=? vp=?

Problem: A proton is placed between two parallel conducting plates in a vacuum as shown. The potential difference between the two plates is 450 V. The proton is released from rest close to the positive plate. What is the kinetic energy of the proton when it reaches the negative plate? Example : Through what potential difference would an electron need to accelerate to achieve a speed of 60% of the speed of light, starting from rest? (The speed of light is 3.00×108 m/s.) The final speed of the electron is vf = 0.6(3.00×108 m/s) = 1.80×108 m/s. At this speed, the energy (non-relativistic) is the kinetic energy, KE =KEf =½mvf² = 0.5(9.11×10-31 kg)(1.80×108 m/s)² = 1.48×10-14 J. This energy must equal the change in potential energy from moving through a potential difference, KEf = ∆PE = -q ∆V. Therefore: ∆V = KE/q = (1.48×10-14 J)/(-1.6×10-19 C) = -9.25×104 V.

Electric potential and potential energy due to point charges
Electric circuits: point of zero potential is defined by grounding some point in the circuit Electric potential due to a point charge at a point in space: point of zero potential is taken at an infinite distance from the charge With this choice, a potential can be found as Note: the potential depends only on charge of an object, q, and a distance from this object to a point in space, r.

Superposition principle for potentials
If more than one point charge is present, their electric potential can be found by applying superposition principle The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges. Remember that potentials are scalar quantities!

Potential energy of a system of point charges
Consider a system of two particles If V1 is the electric potential due to charge q1 at a point P, then work required to bring the charge q2 from infinity to P without acceleration is q2V1. If a distance between P and q1 is r, then by definition Potential energy is positive if charges are of the same sign. q2 q1 r P A

Example: potential energy of an ion
Three ions, Na+, Na+, and Cl-, located such, that they form corners of an equilateral triangle of side 2 nm in water. What is the electric potential energy of one of the Na+ ions? Cl- ? Na+ Na+

Potentials and charged conductors
Recall that work is opposite of the change in potential energy, No work is required to move a charge between two points that are at the same potential. That is, W=0 if VB=VA Recall: all charge of the charged conductor is located on its surface electric field, E, is always perpendicular to its surface, i.e. no work is done if charges are moved along the surface Thus: potential is constant everywhere on the surface of a charged conductor in equilibrium … but that’s not all!

Because the electric field is zero inside the conductor, no work is required to move charges between any two points, i.e. If work is zero, any two points inside the conductor have the same potential, i.e. potential is constant everywhere inside a conductor Finally, since one of the points can be arbitrarily close to the surface of the conductor, the electric potential is constant everywhere inside a conductor and equal to its value at the surface! Note that the potential inside a conductor is not necessarily zero, even though the interior electric field is always zero!

The electron volt A unit of energy commonly used in atomic,
nuclear and particle physics is electron volt (eV) The electron volt is defined as the energy that electron (or proton) gains when accelerating through a potential difference of 1 V Relation to SI: 1 eV = 1.60×10-19 C·V = 1.60×10-19 J Vab=1 V

Example : ionization energy of the electron in a hydrogen atom
In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29×10-11 m. Find the ionization energy of the atom, i.e. the energy required to remove the electron from the atom. Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not a very good model of the atom. A better picture is one in which the electron is spread out around the nucleus in a cloud of varying density; however, the Bohr model does give the right answer for the ionization energy

In the Bohr model of a hydrogen atom, the electron, if it is in the ground state, orbits the proton at a distance of r = 5.29 x m. Find the ionization energy, i.e. the energy required to remove the electron from the atom. Given: r = x m me = 9.11×10-31 kg mp = 1.67×10-27 kg |e| = 1.60×10-19 C Find: E=? The ionization energy equals to the total energy of the electron-proton system, with The velocity of e can be found by analyzing the force on the electron. This force is the Coulomb force; because the electron travels in a circular orbit, the acceleration will be the centripetal acceleration: or or Thus, total energy is

Calculating the Potential from the Electric Field
To calculate the electric potential from the electric field we start with the definition of the work dW done on a particle with charge q by a force F over a displacement ds In this case the force is provided by the electric field F = qE Integrating the work done by the electric force on the particle as it moves in the electric field from some initial point i to some final point f we obtain

Remembering the relation between the change
in electric potential and the work done … …we find Taking the convention that the electric potential is zero at infinity we can express the electric potential in terms of the electric field as

Example - Charge moves in E field
Given the uniform electric field E, find the potential difference Vf-Vi by moving a test charge q0 along the path icf. Idea: Integrate E  ds along the path connecting ic then cf. (Imagine that we move a test charge q0 from i to c and then from c to f.)

Example - Charge moves in E field

Question We just derived Vf-Vi for the path i → c → f. What is Vf-Vi when going directly from i to f ? A: 0 B: -Ed C: +Ed D: -1/2 Ed Quick: DV is independent of path. Explicit: DV = -  E . ds =  E ds = - Ed

(The Electric Potential)
Lecture 8 Application (The Electric Potential)

[1] What potential difference is needed to stop an electron with an initial speed of 4.2×105m/s?
[2] An ion accelerated through a potential difference of 115V experiences an increase in potential energy of 7.37×10-17J. Calculate the charge on the ion.

[3] An infinite charged sheet has a surface charge density σ of 1
[3] An infinite charged sheet has a surface charge density σ of 1.0×10-7 C/m2. How far apart are the equipotential surfaces whose potentials differ by 5.0 V? [4] At what distance from a point charge of 8µC would the potential equal 3.6×104V?

[5] At a distance r away from a point charge q, the electrical potential is V=400v and the magnitude of the electric field is E=150N/C. Determine the value of q and r.

[6] Calculate the value of the electric potential at point P due to the charge configuration shown in Figure. Use the values q1=5mC, q2=-10mC, a=0.4m, and b=0.5m. By substitute find V

[7] Two large parallel conducting plates are 10cm apart and carry equal and opposite charges on their facing surfaces. An electron placed midway between the two plates experiences a force of 1.6×1015N. What is the potential difference between the plates?

[8] Two point charges are located as shown in, where ql=+4mC, q2=-2mC, a=0.30m, and b=0.90m. Calculate the value of the electrical potential at points P1, and P2. Which point is at the higher potential?

[9] In figure prove that the work required to put four charges together on the corner of a square of radius a is given by

[10] Assume we have a system of three point charges: q1 = +1
[10] Assume we have a system of three point charges: q1 = C q2 = C q3 = C. q1 is located at (0,a) q2 is located at (0,0) q3 is located at (b,0) a = 8.00 m and b = 6.00 m. What is the electric potential at point P located at (b,a)?

The electric potential at point P is given by the sum of the electric potential from the three charges r1 r2 r3

[11] An electron initially has velocity 5 x 105 m/s
[11] An electron initially has velocity 5 x 105 m/s. It is accelerated through a potential of 2 V. What is its final velocity?

[12] A charge of -1x10-8C weighs 1 g
[12] A charge of -1x10-8C weighs 1 g. It is released at rest from point P and moves to point Q. It's velocity at point Q is 1 cm/s. What is the potential difference, VP - VQ?

[13] A proton with a speed of vi = 2
[13] A proton with a speed of vi = 2.0 x 105 m/s enters a region of space where source charges have created an electric potential. What is the proton’s speed after it has moved through a potential difference of ∆V = 100 V?

[14] Potential Due to a Charged Rod
A rod of length L located along the x axis has a uniform linear charge density λ. Find the electric potential at a point P located on the y axis a distance d from the origin. Start with then, So Now, let me give you an example to show how to do that. Let’s assume the rod is lying along the x axis, dx is the length of one small segment, and dq is the charge on that segment. So we have dq=lamda times dx. The magnitude of the electric field at P due to this small segment is The total field at P is the integration from a to l+a. Since lamda equals to Q/l, the electric field can be expressed like that. Now, let’s see two extreme cases: (1) l=>0, means the rod has shrunk to zero size,, equation reduces to the electric field due to a point charge. (2) a>>l, means P is far from the rod, then l in the denominator can be neglected. This is exactly the form we would expect for a point charge.

[15] The Electric Potential of a Charged Ring
Find the potential of a thin uniformly charged ring of radius R and charge Q at point P on the z axis?

[16] The Electric Potential of a Charged Disk
Find the potential V of a thin uniformly charged disk of radius R and charge density s at point P on the z axis?

Lecture 9 The Electric Field & Electric Potential Due to Continuous Charge Distributions

Field & Potential Due to a Continuous Charge Distribution
The electric field and electric potential due to a continuous charge distribution is found by treating charge elements as point charges and then summing via integrating, the electric field vectors and the electric potential produced by all the charge elements.

[1] Electric Field Due to a Charged Rod
A rod of length L has a uniform positive charge per unit length λ and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end.

[2] Infinitely Long Line of Charge
+ + y-components cancel by symmetry + + + + + +

[3] Electric Field on the Z-Axis of a Charged Ring
determine the field at point P on the axis of the ring. perpendicular-components cancel by symmetry + + + + + + +

The charged ring must look like a point source.
When: The charged ring must look like a point source. Note that for z >> R (the radius of the ring), this reduces to a simple Coulomb field.

[4] Electric Field on the Axis of an Uniformly Charged Disk
Using the charged ring result,

Two Important Limiting Cases
Large Charged Plate: Very Far From the Charged Plate:

[5] Potential Due to a Charged Rod
A rod of length L located along the x axis has a uniform linear charge density λ. Find the electric potential at a point P located on the y axis a distance d from the origin. Start with then, So Now, let me give you an example to show how to do that. Let’s assume the rod is lying along the x axis, dx is the length of one small segment, and dq is the charge on that segment. So we have dq=lamda times dx. The magnitude of the electric field at P due to this small segment is The total field at P is the integration from a to l+a. Since lamda equals to Q/l, the electric field can be expressed like that. Now, let’s see two extreme cases: (1) l=>0, means the rod has shrunk to zero size,, equation reduces to the electric field due to a point charge. (2) a>>l, means P is far from the rod, then l in the denominator can be neglected. This is exactly the form we would expect for a point charge.

[6] The Electric Potential of a Charged Ring
Find an expression for the electric potential at a point P located on the perpendicular central axis of a uniformly charged ring of radius a and total charge Q

[7] The Electric Potential of a Charged Disk
Find the potential V of a thin uniformly charged disk of radius R and charge density σ at point P on the z axis?

Lecture 10 Capacitance and capacitors

Capacitors Capacitors are devices that store energy in an electric field. Capacitors are used in many every-day applications Heart defibrillators Camera flash units Capacitors are an essential part of electronics. Capacitors can be micro-sized on computer chips or super-sized for high power circuits such as FM radio transmitters.

Definition of Capacitance
The definition of capacitance is The units of capacitance are coulombs per volt. The unit of capacitance has been given the name farad (abbreviated F) named after British physicist Michael Faraday ( ) A farad is a very large capacitance Typically we deal with F (10-6 F), nF (10-9 F), or pF (10-12 F)

The parallel-plate capacitor
The capacitance of a device depends on the area of the plates and the distance between the plates where A is the area of one of the plates, d is the separation, e0 is a constant (permittivity of free space), e0= 8.85×10-12 C2/N·m2 A +Q d A -Q

the charge on each plate
Example: A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate Solution: Since we are dealing with the parallel-plate capacitor, the capacitance can be found as Given: DV=10,000 V A = 2.00 m2 d = 5.00 mm Find: C=? Q=? Once the capacitance is known, the charge can be found from the definition of a capacitance via charge and potential difference:

Cylindrical Capacitor
Consider a capacitor constructed of two collinear conducting cylinders of length L. The inner cylinder has radius r1 and the outer cylinder has radius r2. Both cylinders have charge per unit length  with the inner cylinder having positive charge and the outer cylinder having negative charge. We will assume an ideal cylindrical capacitor The electric field points radially from the inner cylinder to the outer cylinder. The electric field is zero outside the collinear cylinders.

We apply Gauss’ Law to get the electric field between
the two cylinder using a Gaussian surface with radius r and length L as illustrated by the red lines … which we can rewrite to get an expression for the electric field between the two cylinders

As we did for the parallel plate capacitor, we define the
voltage difference across the two cylinders to be V=V1 – V2. The capacitance of a cylindrical capacitor is

Spherical Capacitor Consider a spherical capacitor formed by two
concentric conducting spheres with radii r1 and r2 Let’s assume that the inner sphere has charge +q and the outer sphere has charge –q. The electric field is perpendicular to the surface of both spheres and points radially outward

To calculate the electric field, we use a Gaussian surface consisting of a concentric sphere of radius r such that r1 < r < r2 The electric field is always perpendicular to the Gaussian surface so … which reduces to

To get the electric potential we follow a method similar to the one we used for the cylindrical capacitor and integrate from the negatively charged sphere to the positively charged sphere Using the definition of capacitance we find The capacitance of a spherical capacitor is then

Combinations of capacitors
It is very often that more than one capacitor is used in an electric circuit We would have to learn how to compute the equivalent capacitance of certain combinations of capacitors C2 C5 C1 C2 C3 C4 C1 C3

a. Parallel combination
Connecting a battery to the parallel combination of capacitors is equivalent to introducing the same potential difference for both capacitors, A total charge transferred to the system from the battery is the sum of charges of the two capacitors, +Q1 -Q1 C1 V=Vab a b +Q2 -Q2 C2

Parallel combination:
Analogous formula is true for any number of capacitors, It follows that the equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors (parallel combination)

First determine equivalent capacitance of C1 and C2:
Example: A 3 mF capacitor and a 6 mF capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. +Q1 -Q1 C1 V=Vab a b +Q2 -Q2 C2 Given: V = 18 V C1= 3 mF C2= 6 mF Find: Ceq=? Q=? First determine equivalent capacitance of C1 and C2: Next, determine the charge

Series combination Connecting a battery to the serial combination of capacitors is equivalent to introducing the same charge for both capacitors, A voltage induced in the system from the battery is the sum of potential differences across the individual capacitors, +Q1 -Q1 C1 +Q2 -Q2 C2 V=Vab a c b

Series combination: Analogous formula is true for any number of
capacitors, It follows that the equivalent capacitance of a series combination of capacitors is always less than any of the individual capacitance in the combination (series combination)

Example: A 3 mF capacitor and a 6 mF capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited. +Q1 -Q1 C1 +Q2 -Q2 C2 V=Vab a c b Given: V = 18 V C1= 3 mF C2= 6 mF Find: Ceq=? Q=? First determine equivalent capacitance of C1 and C2: Next, determine the charge

Example: Capacitors in Series and Parallel
Three capacitors are connected as shown. (a) Find the equivalent capacitance of the 3-capacitor combination. (b) The capacitors, initially uncharged, are connected across a 6.0 V battery. Find the charge and voltage drop for each capacitor.

Energy stored in a charged capacitor
Consider a battery connected to a capacitor A battery must do work to move electrons from one plate to the other. The work done to move a small charge q across a voltage V is W = V q. As the charge increases, V increases so the work to bring q increases. Using calculus we find that the energy (U) stored on a capacitor is given by: V V q Q

Example: electric field energy in parallel-plate capacitor
Find electric field energy density (energy per unit volume) in a parallel-plate capacitor Recall Thus, and so, the energy density is + -

(a) determine the equivalent capacitance of the circuit,
Example: In the circuit shown V = 48V, C1 = 9mF, C2 = 4mF and C3 = 8mF. (a) determine the equivalent capacitance of the circuit, (b) determine the energy stored in the combination by calculating the energy stored in the equivalent capacitance, C1 C2 V C3 The energy stored in the capacitor C123 is then

Capacitors with dielectrics
A dielectrics is an insulating material (rubber, glass, etc.) Consider an insolated, charged capacitor Notice that the potential difference decreases (k = V0/V) Since charge stayed the same (Q=Q0) → capacitance increases dielectric constant: k = C/C0 Dielectric constant is a material property +Q -Q Insert a dielectric V0 V

Capacitance is multiplied by a factor k when the
dielectric fills the region between the plates completely E.g., for a parallel-plate capacitor The capacitance is limited from above by the electric discharge that can occur through the dielectric material separating the plates In other words, there exists a maximum of the electric field, sometimes called dielectric strength, that can be produced in the dielectric before it breaks down

Dielectric constants and dielectric strengths of various materials at room temperature
Dielectric constant, k Dielectric strength (V/m) Vacuum 1.00 -- Air 3 × 106 Water 80 Fused quartz 3.78 9× 106

Example: Take a parallel plate capacitor whose plates have an area of 2 m2 and are separated by a distance of 1cm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kV. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material. Since we are dealing with the parallel-plate capacitor, the original capacitance can be found as The dielectric constant and the new capacitance are The charge on the capacitor can be found to be

How does an insulating dielectric material reduce electric fields by
producing effective surface charge densities? Reorientation of polar molecules + - + - Induced polarization of non-polar molecules + - Dielectric Breakdown: breaking of molecular bonds/ionization of molecules.

Solved Problems (Electrostatic)

[1] Three charges are located in the XY plane as Shown in the figure
[1] Three charges are located in the XY plane as Shown in the figure. Q1= - 3 nC, Q2 = 5 nC Q3 = 3 nC. The distance between Q1 and Q2 is 8 m and between Q2 and Q3 is 6 m. Find: The electric field due to the charges at the origin O. The electric potential due to these charges at the origin O. c) What is the direction of the net force on the point O.

The electric field due to the charges at the origin O.

b) The electric potential due to these charges at the origin O.
c) What is the direction of the net force on the point O.

[2] Two point charges are located on the x-axis as follows: charge Q1 = 8x10-9 C is at x = 0 and charge Q2 = -1x10-9 C is at x = 1cm. What is the net force in the x direction on a third charge,Q3 = +2x10-9 C, placed at x = 2 cm ?

[3] In a hydrogen atom an electron circles a proton
[3] In a hydrogen atom an electron circles a proton. Since the mass of the proton is much greater than the mass an electron, assume that the proton stays fixed in space and the electron rotates around it with radius 0.529x10-10 m. What is the velocity of the electron?

[4] A uniform electric field exists in the region between two oppositely charged plane parallel plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate 2x10-8 s later. If the magnitude of the electric field is 4x103 N/C, what is the separation between the plates?

[5] Two charged particles are placed on the x-axis as follows: Q1 = 1x10-9 C at x = 0 m and Q2 = -2x10-9 C at x = 1.6 m. Where on the x-axis is the electric field zero? The electric field can only be zero where the field due to Q1 and Q2 point in opposite direction. This occurs both for x<0m and for x>1.6 m, however since Q2 is greater in magnitude, one must be further away from it in order to have the magnitudes of the fields due to Q1 and Q2 be equal. This means the field is zero in the region x<0. Sitting the magnitude equal,

[6] Two stationary point charges are arranged on the x-axis as follows: Q1 = 2.4x10-10 C at x = 0 m and Q2 = -1.2x10-10 C at x = 1 m. An electron is placed at x = 2 m and let go. What is the velocity of the electron when it reaches x = 3 m. This is a conservation of energy problem. The initial kinetic energy of the electron is zero. The initial and final potential energy of the electron are: Consequently, the final kinetic energy is And the final velocity of the electron is

[7] Using the superposition principle determine the electric field at point P in Fig.
Using the convention that a positive electric field is up and a negative electric field is down, the net electric field at p is

[8] A 1 Coulomb charge is located at the origin
[8] A 1 Coulomb charge is located at the origin. Another charge, Q, is located at x = 3 m. If the electric field is zero at x = 1 m, what is Q? The electric field due to the 1C charge is in positive x-direction at x=1m. In order to cancel this field, the charge at x=3m must be positive so that its electric field is in the negative x-direction at x=1m. The two field will cancel if they have the same magnitude.

a) Find the equivalent capacitance of the entire combination.
[9] a) Find the equivalent capacitance of the entire combination. V C1 C2 +- C3 C1 and C2 are in series. C1 = 10 F C2 = 5.0 F C3 = 4.0 F C12 and C3 are in parallel.

b) If V = 100 volts, what is the charge Q3 on C3?
C = Q/V c) What is the total energy stored in the circuit?

[10] A wire of uniform charge density and length L lies along the x
axis as shown in Figure. What is the electric potential at point A?

[11] Determine the point nearest the charges where the total electric field is zero. Take the µC charge to be at the origin of the x axis. 1.00 m 0.39 m -2.50 C -6.00 C [12] The plates of a parallel plate capacitor are separated by 1 x 10–4 m. If the material in between them is a jelly with a dielectric constant of 2.26, what is the plate area needed to provide a capacitance of 1.50 pF? 7.50 x 10-6 m2

[13] Find the charge on the 2 F capacitor in the following circuit.
12 V [14] Each of the protons in a particle beam has a kinetic energy of 3.25x10-15 J What are the magnitude and direction of the electric field that will stop these protons in a distance of 1.25 m? 1.63 x 104 N/C and opposite to the motion

[15] Three point charges are fixed on the corners of an equilateral triangle whose one side is b as shown in Figure. 1. What is the magnitude of the Coulomb force acting on charge –q due to presence of other charges? 2. What is the value of the electric potential at the center (point A) of positive charges? 3. What is the electric potential energy of system?

[16] The stored energy of a capacitor is 3
[16] The stored energy of a capacitor is 3.0 J after having been charged by a 1.5 V battery. What is the energy of the capacitor after it is charged by 3.0 V battery? [17] Find the equivalent capacitance between points a and b in the combination of capacitors shown in Figure. 7µF and 5µF series 4µF, 6µF and 2.9µF parallel

[18] Two capacitors when connected in parallel give an equivalent capacitance of 9.00 pF and give an equivalent capacitance of 2.00 pF when connected in series. What is the capacitance of each capacitor? in parallel in series Solve eq.1&2

Lecture 12 Current & Resistance

Electric Current Definition: the current is the rate at which charge flows through this surface. Given an amount of charge, DQ, passing through the area A in a time interval Dt, the current is the ratio of the charge to the time interval. The SI units of current is the ampere (A). 1 A = 1 C/s 1 A of current is equivalent to 1 C of charge passing through the area in a time interval of 1 s.

Example: The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find the current in the light bulb. Find no. of electrons?

Current Density When we care only about the total current I in a conductor, we do not have to worry about its shape. However, sometimes we want to look in more detail at the current flow inside the conductor. Similar to what we did with Gauss’ Law (electric flux through a surface), we can consider the flow of charge through a surface. To do this, we consider (charge per unit time) per unit area, i.e. current per unit area, or current density. The units are amps/square meter (A/m2). Current density is a vector (since it has a flow magnitude and direction). We use the symbol The relationship between current and current density is High current density Small current density

Current and Drift Speed
Consider the current on a conductor of cross-sectional area A.

Volume of an element of length Dx is : DV = A Dx.
Let n be the number of carriers per unit of volume. The total number of carriers in DV is: n A Dx. The charge in this volume is: DQ = (n A Dx)q. Distance traveled at drift speed vd by carrier in time Dt: Dx = vd Dt. Hence: DQ = (n A vd Dt)q. The current through the conductor: I = DQ/ Dt = n A vd q. The current density : J = I/A = n vd q.

Example: A copper wire of cross-sectional area 3.00x10-6 m2 carries a current of 10 A. Assuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. A = 3.00x10-6 m2 ; I = 10 A, q = 1.6 x C. n = 8.48 x 1022 electrons/ m3. [Q] If the current density in a copper wire is equal to 5.8×106A/m2, calculate the drift velocity of the free electrons in this wire.

Drift speeds are usually very small.
Drift speed much smaller than the average speed between collisions. Electrons traveling at 2.46x10-6 m/s would take 68 min to travel 1m. So why does light turn on so quickly when one flips a switch? The info (electric field) travels at roughly 108 m/s… [Q] A silver wire 1 mm in diameter transfers a charge of 65 C in 1 hr, 15 min. Silver contains 5.80 x 1028 free electrons per cubic meter. a) What is the current in the wire? b) What is the magnitude of the drift velocity of the electrons in the wire? Ans. a) A; b) 1.98 x 10-6 m/s

Resistance and Ohm’s Law
When a voltage (potential difference) is applied across the ends of a metallic conductor, the current is found to be proportional to the applied voltage. In situations where the proportionality is exact, one can write. The proportionality constant R is called resistance of the conductor.

The resistance is defined as the ratio.
In SI, resistance is expressed in volts per ampere. A special name is given: ohms Example: if a potential difference of 10 V applied across a conductor produces a 0.2 A current, then one concludes the conductors has a resistance of 10 V/0.2 A = 50 W.

Ohm’s Law Resistance in a conductor arises because of collisions between electrons and fixed charges within the material. In many materials, including most metals, the resistance is constant over a wide range of applied voltages. This is a statement of Ohm’s law. Ohm’s Law

The current–potential difference curve for an ohmic material
The current–potential difference curve for an ohmic material. The curve is linear, and the slope is equal to the inverse of the resistance of the conductor. A nonlinear current–potential difference curve for a junction diode. This device does not obey Ohm’s law.

Resistivity Electrons moving inside a conductor subject to an external potential constantly collide with atoms of the conductor. They lose energy and are repeated re-accelerated by the electric field produced by the external potential. The collision process is equivalent to an internal friction. This is the origin of a material’s resistance.

The resistance of an ohmic conductor is proportional to the its length, l, and inversely proportional to the cross section area, A, of the conductor. The constant of proportionality r is called the resistivity of the material. Every material has a characteristic resistivity that depends on its electronic structure, and the temperature. Good conductors have low resistivity. Insulators have high resistivity.

Resistivity - Units Resistance expressed in Ohms, Length in meter.
Area are m2, Resistivity thus has units of W m.

Resistivity of various materials
Resistivity (10-8 Wm) Silver 1.61 Bismuth 106.8 Copper 1.70 Plutonium 141.4 Gold 2.20 Graphite 1375 Aluminum 2.65 Germanium 4.6x107 Pure Silicon 3.5 Diamond 2.7x109 Calcium 3.91 Deionized water 1.8x1013 Sodium 4.75 Iodine 1.3x1015 Tungsten 5.3 Phosphorus 1x1017 Brass 7.0 Quartz 1x1021 Uranium 30.0 Alumina 1x1022 Mercury 98.4 Sulfur 2x1023

Example (a) Calculate the resistance per unit length of a nichrome wire of radius m. Cross section: Resistivity (Table): 1.5 x 10-6 Wm. Resistance/unit length: (b) If a potential difference of 10.0 V is maintained across a 1.0-m length of the nichrome wire, what is the current?

The reciprocal of the resistivity is called the conductivity,
[Q] Stretching changes resistance: A wire of resistance R is stretched uniformly until it is twice its original length. What happens to its resistance? The resistance of the wire increases by a factor of four if the length increases twice [Q] Speaker wires: Suppose you want to connect your stereo to remote speakers. (a) If each wire must be 20m long, what diameter copper wire should you use to keep the resistance less than 0.1Ω per wire? (b) If the current on each speaker is 4.0A, what is the voltage drop across each wire? [Q] A 2.4m length of wire that is 0.031cm2 in cross section has a measured resistance of 0.24Ω. Calculate the conductivity of the material.

Temperature Variation of Resistance
The resistivity of a metal depends on many (environmental) factors. The most important factor is the temperature. For most metals, the resistivity increases with increasing temperature. The increased resistivity arises because of larger friction caused by the more violent motion of the atoms of the metal.

For most metals, resistivity increases approx
For most metals, resistivity increases approx. linearly with temperature. r is the resistivity at temperature T (measured in Celsius). ro is the reference resistivity at the reference temperature To (usually taken to be 20 oC). a is a parameter called temperature coefficient of resistivity. For a conductor with fixed cross section.

Example: A resistance thermometer, which measures temperature by measuring the change in the resistance of a conductor, is made of platinum and has a resistance of 50.0 W at 20oC. When the device is immersed in a vessel containing melting indium, its resistance increases to W. Find the melting point of Indium. Using a=3.92x10-3(oC)-1 from table. Ro=50.0 W. To=20oC. R=76.8 W.

[Q] A resistance thermometer using a platinum wire is used to measure the temperature of a liquid. The resistance is 2.42 ohms at 0oC, and when immersed in the liquid it is 2.98 ohms. The temperature coefficient of resistivity of platinum is What is the temperature of the liquid?

Superconductivity 1911: H. K. Onnes, who had figured out how to make liquid helium, used it to cool mercury to 4.2 K and looked at its resistance At low temperatures the resistance of some metals0, measured to be less than 10-16•ρconductor (i.e., ρ<10-24 Ωm)! Resistance versus temperature for a sample of mercury (Hg). The graph follows that of a normal metal above the critical temperature Tc. The resistance drops to zero at Tc, which is 4.2 K for mercury.

Electrical energy and power
In any circuit, battery is used to induce electrical current chemical energy of the battery is transformed into kinetic energy of mobile charge carriers (electrical energy gain) Any device that possesses resistance (resistor) present in the circuit will transform electrical energy into heat kinetic energy of charge carriers is transformed into heat via collisions with atoms in a conductor (electrical energy loss)

Electrical energy Consider circuit on the right in detail
AB: charge gains electrical energy form the battery (battery looses chemical energy) CD: electrical energy lost (transferred into heat) Back to A: same potential energy (zero) as before Gained electrical energy = lost electrical energy on the resistor

Power Compute rate of energy loss (power dissipated on the resistor)
Use Ohm’s law Units of power : watt delivered energy: kilowatt-hours

Example A high-voltage transmission line with resistance of 0.31 W/km carries 1000A , starting at 700 kV, for a distance of 160 km. What is the power loss due to resistance in the wire? Observations: Given resistance/length, compute total resistance Given resistance and current, compute power loss Now compute power

Lecture 13 Direct Current Circuits

A current is maintained in a closed circuit by a source of emf.
What is electromotive force (emf)? A current is maintained in a closed circuit by a source of emf. The term emf was originally an abbreviation for electromotive force but emf is NOT really a force, so the long term is discouraged. A source of emf works as “charge pump” that forces electrons to move in a direction opposite the electrostatic field inside the source. Examples of such sources are: batteries generators thermocouples photo-voltaic cells

Each real battery has some internal resistance
AB: potential increases by on the source of EMF, then decreases by Ir (because of the internal resistance) Thus, terminal voltage on the battery DV is Note: EMF is the same as the terminal voltage when the current is zero (open circuit)

Now add a load resistance R
Since it is connected by a conducting wire to the battery → terminal voltage is the same as the potential difference across the load resistance Thus, the current in the circuit is [Q] Under what condition does the potential difference across the terminals of a battery equal its emf?

Resistors in series 1. Because of the charge conservation, all charges going through the resistor R2 will also go through resistor R1. Thus, currents in R1 and R2 are the same, 2. Because of the energy conservation, total potential drop (between A and C) equals to the sum of potential drops between A and B and B and C, By definition, Thus, Req would be

Analogous formula is true for any number of resistors,
It follows that the equivalent resistance of a series combination of resistors is greater than any of the individual resistors (series combination) [Q] How would you connect resistors so that the equivalent resistance is larger than the individual resistance? [Q] When resistors are connected in series, which of the following would be the same for each resistor: potential difference, current, power?

Example In the electrical circuit below, find voltage across the resistor R1 in terms of the resistances R1, R2 and potential difference between the battery’s terminals V. Energy conservation implies: with Then, Thus, This circuit is known as voltage divider.

Resistors in parallel 1. Since both R1 and R2 are connected to the same battery, potential differences across R1 and R2 are the same, 2. Because of the charge conservation, current, entering the junction A, must equal the current leaving this junction, By definition, Thus, Req would be or

Analogous formula is true for any number of resistors,
It follows that the equivalent resistance of a parallel combination of resistors is always less than any of the individual resistors (parallel combination) [Q] How would you connect resistors so that the equivalent resistance is smaller than the individual resistance? [Q] When resistors are connected in parallel, which of the following would be the same for each resistor: potential difference, current, power?

example In the electrical circuit below, find current through the resistor R1 in terms of the resistances R1, R2 and total current I induced by the battery. Charge conservation implies: with Then, Thus, This circuit is known as current divider.

Example Find the currents in the circuit shown

Example Find the currents I1 and I2 and the voltage Vx in the circuit shown below. First find the equivalent resistance seen by the 20 V source: We now find I1 and I2 directly from the current division rule: Finally, voltage Vx is

Kirchhoff’s Rules The sum of currents entering any junction must equal the sum of the currents leaving that junction (current or junction rule) . The sum of the potential differences across all the elements around any closed-circuit loop must be zero (voltage or loop rule). Charge conservation Energy conservation

Rules for Kirchhoff’s loop rule

Solving problems using Kirchhoff’s rules

Example

Example Find all three currents
Need three equations for three unknowns Note that current directions are already picked for us (sometimes have to pick for yourself) Use the junction rule first Alternative two loops

RC Circuits A direct current circuit may contain capacitors and resistors, the current will vary with time When the circuit is completed, the capacitor starts to charge The capacitor continues to charge until it reaches its maximum charge (Q = Cξ) Once the capacitor is fully charged, the current in the circuit is zero

Charging Capacitor in an RC Circuit
The charge on the capacitor varies with time q = Q(1 – e-t/RC) The time constant, =RC The current I is The time constant represents the time required for the charge to increase from zero to 63.2% of its maximum

Notes on Time Constant In a circuit with a large time constant, the capacitor charges very slowly The capacitor charges very quickly if there is a small time constant After t = 10 t, the capacitor is over 99.99% charged

Discharging Capacitor in an RC Circuit
When a charged capacitor is placed in the circuit, it can be discharged q = Qe-t/RC The charge decreases exponentially The current I is At t =  = RC, the charge decreases to Qmax In other words, in one time constant, the capacitor loses 63.2% of its initial charge

Example : charging the unknown capacitor
A series combination of a 12 kW resistor and an unknown capacitor is connected to a 12 V battery. One second after the circuit is completed, the voltage across the capacitor is 10 V. Determine the capacitance of the capacitor. I R C

Recall that the charge is building up according to
Thus the voltage across the capacitor changes as This is also true for voltage at t = 1s after the switch is closed,

[1] What is the magnitude of the current flowing in the circuit shown in Fig. ? The net voltage drop due to the batteries 0V so no current flows. I=0A [2] A copper wire has resistance 5 Ohms. Given that the resistivity of silver is 85 percent of the resistivity of copper, what is the resistance of a silver wire three times as long with twice the diameter?

[3] A resistor draws a current of 1A when connected across an ideal 3V battery. Another resistor draws a current of 2A when connected across an ideal 3V battery. What current do the two resistors draw when they are connected in series across an ideal 3V battery? The first resistor has a resistance of The second resistor has a resistance of The series combination of the two resistors is Which when connected across a 3V battery will draw a current of

[4] consider an RC circuit in which the capacitor is being by a battery connected in the circuit. In five time constant, what percentage of final charge is on the capacitor?

[4] In fig. (a) find the time constant of the circuit and the charge in the capacitor after the switch is closed. (b) find the current in the resistor R at time 10 sec after the switch is closed. Assume R=1×106 Ω, emf =30 V and C=5×10-6F

[5] In the circuit shown in Fig. what is the current labeled I?

[6] A certain wire has resistance R
[6] A certain wire has resistance R. What is the resistance of a second wire, made of the same material, which is half as long and has 1/3 the diameter? The resistance is proportional to the length of the wire and inversely proportional to the area. Since area is proportional to the diameter squared, the resistance is

[7] The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time according to the equation q =4t3 +5t +6, where t is in seconds. (a) What is the instantaneous current through the surface at t = 1.00 s ? (b) What is the value of the current density?

[8] An electric current is given by the expression I(t) =100 sin(120πt), where I is in amperes and t is in seconds. What is the total charge carried by the current from t = 0 to t= (1/240) s?

[9] Suppose that you wish to fabricate a uniform wire out of 1
[9] Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to have a resistance of R = Ω, and if all of the copper is to be used, what will be (a) the length and (b) the diameter of this wire?

[10] Compute the cost per day of operating a lamp that draws a current of 1.70 A from a 110V line. Assume the cost of energy from the power company is $ /kWh.

Lecture 15 Magnetic Fields & Force

Magnets ... two poles: N and S
Like poles repel Unlike poles attract

PERMANENT MAGNETS Figure (a) Magnetic field pattern surrounding a bar magnet as displayed with iron filings. (b) Magnetic field pattern between unlike poles of two bar magnets. (c) Magnetic field pattern between like poles of two bar magnets

Magnetic Field lines: (defined in same way as electric field lines, direction and density)

Broken Permanent Magnet
If we break a permanent magnet in half, we do not get a separate north pole and south pole. When we break a bar magnet in half, we always get two new magnets, each with its own north and south pole.

Source of Magnetic Fields
What is the source of magnetic fields? Answer: electric charge in motion e.g., current in wire surrounding cylinder (solenoid) produces very similar field to that of magnets. Therefore, understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter. Orbits of electrons about nuclei Intrinsic “spin” of electrons (more important effect)

Paramagnetic (aluminum, tungsten, oxygen,…)
Magnetic Materials Materials can be classified by how they respond to an applied magnetic field, Bapp. Paramagnetic (aluminum, tungsten, oxygen,…) Atomic magnetic dipoles (~atomic bar magnets) tend to line up with the field, increasing it. But thermal motion randomizes their directions, so only a small effect persists: Bind ~ Bapp •10-5 Diamagnetic (gold, copper, water,…) The applied field induces an opposing field; again, this is usually very weak; Bind ~ -Bapp •10-5 [Exception: Superconductors exhibit perfect diamagnetism  they exclude all magnetic fields] Ferromagnetic (iron, cobalt, nickel,…) Somewhat like paramagnetic, the dipoles prefer to line up with the applied field. But there is a complicated collective effect due to strong interactions between neighboring dipoles  they tend to all line up the same way. Very strong enhancement. Bind ~ Bapp •10+5

Magnetic Field Direction
A vector quantity: magnitude and direction… The letter B is used to represent magnetic fields.

Magnetic Field of the Earth
A small magnetic bar should be said to have north and south seeking poles. The north of the bar points towards the North of the Earth. The geographic north corresponds to a south magnetic pole and the geographic south corresponds to a magnetic north. The configuration of the Earth magnetic resemble that of a (big) magnetic bar one would put in its center.

Magnetic Field of the Earth

Magnetic Fields in analogy with Electric Fields
Distribution of charge creates an electric field E in the surrounding space. Field exerts a force F=q E on a charge q Magnetic Field: Moving charge or current creates a magnetic field B in the surrounding space. Field exerts a force F on a charge moving q

The magnetic force Observations show that the force is proportional to
The field The charge The velocity of the particle The sine of the angle between the field and the direction of the particle’s motion.

Strength and direction of the Magnetic Force on a charge in motion
[Q] The figure shows a charged particle with velocity v travels through a uniform magnetic field B. What is the direction of the magnetic force F on the particle?

Magnetic Field Magnitude
[Q] An electron (q = -1.6x10-19 C) is moving at 3 x 105 m/s in the positive x direction. A magnetic field of 0.8T is in the positive z direction. The magnetic force on the electron is:

Magnetic Field Units [F] = Newton [v] = m/s [q] = C [B] = tesla (T).
Also called weber (Wb) per square meter. 1 T = 1 Wb/m2. 1 T = 1 N s m-1 C-1. 1 T = 1 N A-1 m-1. CGS unit is the Gauss (G) 1 T = 104 G.

Right Hand Rule

Magnetic Force on Current- carrying conductor.
A magnetic force is exerted on a single charge in motion through a magnetic field. That implies a force should also be exerted on a collection of charges in motion through a conductor I.e. a current. The force on a current is the sum of all elementary forces exerted on all charge carriers in motion.

Magnetic Force on Current
If B is directed into the page we use blue crosses representing the tail of arrows indicating the direction of the field, If B is directed out of the page, we use dots. If B is in the page, we use lines with arrow heads.

Force on a wire carrying current in a magnetic field.

Force on a wire carrying current in a magnetic field.
General Case: field at angle q relative to current. B B sin q q I Note: If wire is not straight, compute force on differential elements and integrate:

Example: Wire in Earth’s B Field
A wire carries a current of 22 A from east to west. Assume that at this location the magnetic field of the earth is horizontal and directed from south to north, and has a magnitude of 0.50 x 10-4 T. Find the magnetic force on a 36-m length of wire. What happens if the direction of the current is reversed?

For the straight portion: F1=IB2R out of the page
For the curved portion: If q is the angle between B and ds, then the magnitude of dF2 is dF2= IdSXB= I B sinq ds Because s=Rq we have ds=Rdq  dF2= I B R sinq dq The resultant force F2 on the curved wire must be into the page. Integrating our expression for dF2 over the limits q=0 to q=p (that is, the entire semicircle) gives The net force acting on a closed current loop in a uniform magnetic field is zero.

Magnetic field out of page. What is net force on wire?
Example: Wire with current i. Magnetic field out of page. What is net force on wire? By symmetry, F2 will only have a vertical component,

Lecture 16 Magnetic Fields& Force

Torque on a Current Loop
Imagine a current loop in a magnetic field as follows: B I b a a/2 F In a motor, one has “N” loops of current

Example: A circular loop of radius 50.0 cm is oriented at an angle of 30.0o to a magnetic field of 0.50 T. The current in the loop is 2.0 A. Find the magnitude of the torque. 30.0o B

Galvanometer Device used in the construction of ammeters and voltmeters.

Galvanometer used as Ammeter
Typical galvanometer have an internal resistance of the order of 60 W - that could significantly disturb (reduce) a current measurement. Built to have full scale for small current ~ 1 mA or less. Must therefore be mounted in parallel with a small resistor or shunt resistor. Galvanometer 60 W Rp

Galvanometer used as Voltmeter
Finite internal resistance of a galvanometer must also addressed if one wishes to use it as voltmeter. Must mounted a large resistor in series to limit the current going though the voltmeter to 1 mA. Must also have a large resistance to avoid disturbing circuit when measured in parallel. Galvanometer 60 W Rs

Motion of Charged Particle in magnetic field
Consider positively charge particle moving in a uniform magnetic field. Suppose the initial velocity of the particle is perpendicular to the direction of the field. Then a magnetic force will be exerted on the particle and make follow a circular path.

The magnetic force produces a centripetal acceleration.
The particle travels on a circular trajectory with a radius: What is the period of revolution of the motion? The frequency

Example : Proton moving in uniform magnetic field
A proton is moving in a circular orbit of radius 14 cm in a uniform magnetic field of magnitude 0.35 T, directed perpendicular to the velocity of the proton. Find the orbital speed of the proton.

Example: If a proton moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what is the speed of the proton and the frequency of motion? v r x

Example: Mass Spectrometer
Suppose that B=80mT, V=1000V. A charged ion ( C) enters the chamber and strikes the detector at a distance x=1.6254m. What is the mass of the ion? Key Idea: The uniform magnetic field forces the ion on a circular path and the ion’s mass can be related to the radius of the circular trajectory.

Cyclotrons A cyclotron is a particle accelerator
The D-shaped pieces (descriptively called “dees”) have alternating electric potentials applied to them such that a positively charged particle always sees a negatively charged dee ahead when it emerges from under the previous dee, which is now positively charged

The resulting electric field accelerates the particle
Because the cyclotron sits in a strong magnetic field, the trajectory is curved The radius of the trajectory is proportional to the momentum, so the accelerated particle spirals outward

Example: Deuteron in Cyclotron
Suppose a cyclotron is operated at frequency f=12 MHz and has a dee radius of R=53cm. What is the magnitude of the magnetic field needed for deuterons to be accelerated in the cyclotron (m= kg)?

Example A) 0.9 10-14 J B) 8.47 10-13 J C) 2.7 10-12 J D) 3.74 10-13 J
Suppose a cyclotron is operated at frequency f=12 MHz and has a dee radius of R=53cm. What is the kinetic energy of the deuterons in this cyclotron when they travel on a circular trajectory with radius R (m= kg, B=1.57 T)? A) J B) J C) J D) J

Lecture 17 Sources of the Magnetic Field

History 1819 Hans Christian Oersted discovered that a compass needle was deflected by a current carrying wire Then in 1920s Jean-Baptiste Biot and Felix Savart performed experiments to determine the force exerted on a compass by a current carrying wire

Biot & Savart’s Results
dB the magnetic field produced by a small section of wire ds a vector the length of the small section of wire in the direction of the current r the positional vector from the section of wire to where the magnetic field is measured I the current in the wire  angle between ds & r

dB perpendicular to ds |dB| inversely proportional to |r|2 |dB| proportional to current I |dB| proportional to |ds| |dB| proportional to sin q

Biot–Savart Law All these results could be summarised by one “Law”
Putting in the constant Where m0 is the permeablity of free space

Magnetic Field due to Currents
The passage of a steady current in a wire produces a magnetic field around the wire. Field form concentric lines around the wire Direction of the field given by the right hand rule. If the wire is grasped in the right hand with the thumb in the direction of the current, the fingers will curl in the direction of the field. Magnitude of the field

Magnetic Field of a current loop
Magnetic field produced by a wire can be enhanced by having the wire in a loop. Dx1 1 loop Current  I B I N loops Current  NI Dx2

Ampere’s Law Consider a circular path surrounding a current, divided in segments ds, Ampere showed that the sum of the products of the field by the length of the segment is equal to mo times the current.

Example By way of illustration, let us use Ampere's law to find the magnetic field at a distance r from a long straight wire, a problem we have solved already using the Biot-Savart law

Now consider the interior of the wire, where r < R
Now consider the interior of the wire, where r < R. Here the current i passing through the plane of circle 2 is less than the total current I. Because the current is uniform over the cross section of the wire, the fraction of the current enclosed by circle 2 must equal the ratio of the area πr2 enclosed by circle 2 to the cross-sectional area πR2 of the wire

Magnetic Force between two parallel conductors

Definition of the SI unit Ampere
Used to define the SI unit of current called Ampere. If two long, parallel wires 1 m apart carry the same current, and the magnetic force per unit length on each wire is 2x10-7 N/m, then the current is defined to be 1 A.

Example Two wires, each having a weight per units length of 1.0x10-4 N/m, are strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 mm what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current). l 1 I1 2 d I2

F1 1 I1 B2 mg/l 2 d I2 l

Magnetic Field of a solenoid
Solenoid magnet consists of a wire coil with multiple loops. It is often called an electromagnet.

Solenoid Magnet Field lines inside a solenoid magnet are parallel, uniformly spaced and close together. The field inside is uniform and strong. The field outside is non uniform and much weaker. One end of the solenoid acts as a north pole, the other as a south pole. For a long and tightly looped solenoid, the field inside has a value:

Since the field lines are straight inside the solenoid, the best choice for amperian loop is a rectangle: abcd. Winding density: n=N/L where N = total number of windings and L = total length. Integrate:

Solenoid Magnet The field inside has a value:
n = N/L : number of (loop) turns per unit length. I : current in the solenoid.

Example: Consider a solenoid consisting of 100 turns of wire and length of 10.0 cm. Find the magnetic field inside when it carries a current of A.

Lecture 18 Applications

[1] What is the net force on the rectangular loop of wire in Fig. ?
The force on the top and bottom of the loop are equal and opposite so they cancel.

[2] What is the net force per unit length on the wire carrying 2A in Fig. ?
The 2A wire is attracted to the 1A wire (force up) and attracted to the 3A wire ( force down). Thus, the force per unit length on the 2A wire is

[3] What is the magnitude and direction
of the magnetic field at point P in Fig. ?

[4] What is the magnitude and direction of the magnetic field at point Q in Fig.?

[5] An electron moves is a circular orbit with diameter 10 cm in a 5 Tesla magnetic field. What is the time it takes the electron to complete one orbit?

[6] A proton moves with a velocity of v = (2iˆ- 4jˆ +kˆ ) m/s in
a region in which the magnetic field is B " (iˆ + 2jˆ - 3kˆ ) T. What is the magnitude of the magnetic force this charge experiences?

[7] A conductor suspended by two flexible wires as shown in Figure has a mass per unit length of kg/m. What current must exist in the conductor in order for the tension in the supporting wires to be zero when the magnetic field is 3.60 T into the page? What is the required direction for the current?

[8] A positive charge q = 3.20 × 10-19 C moves with a velocity
v = (2iˆ + 3jˆ - kˆ ) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving charge (in unit-vector notation), taking B = (2iˆ + 4jˆ + kˆ ) T and E =(4iˆ - jˆ - 2kˆ ) V/m. (b) What angle does the force vector make with the positive x axis?

[9] A nonconducting sphere has mass 80. 0 g and radius 20. 0 cm
[9] A nonconducting sphere has mass 80.0 g and radius 20.0 cm. A flat compact coil of wire with 5 turns is wrapped tightly around it, with each turn concentric with the sphere. As shown in Figure, the sphere is placed on an inclined plane that slopes downward to the left, making an angle θ with the horizontal, so that the coil is parallel to the inclined plane. A uniform magnetic field of T vertically upward exists in the region of the sphere. What current in the coil will enable the sphere to rest in equilibrium on the inclined plane? Show that the result does not depend on the value of !.

[10] The segment of wire in Figure carries a current of I =5
[10] The segment of wire in Figure carries a current of I =5.00 A, where the radius of the circular arc is R = 3.00 cm. Determine the magnitude and direction of the magnetic field at the origin.

[11]Figure shows a section of a long hallow cylindrical of radii a=5cm and b=10 cm, carrying a uniform distributed current I, the magnitude of the magnetic field on its outer surface at r=b is measured to be B=0.2T, where r is the radial distance from the cylindrical axis. (a) Find the current in the wire?

(b) Find the magnitude of the magnetic field at r=20 cm?

(c) Find the magnitude of the magnetic field at r=8 cm?

Lecture 19 Faraday’s Law

magnetic flux Definition of Magnetic Flux The flux, F, is defined as the product of the field magnitude by the area crossed by the field lines. where is the component of B perpendicular to the loop, q is the angle between B and the normal to the loop. Units: T·m2 or Webers (Wb)

The flux through the plane is zero when the magnetic field is parallel to the plane surface.
The flux through the plane is a maximum when the magnetic field is perpendicular to the plane.

Example: A square loop 2.00m on a side is placed in a magnetic field of strength 0.300T. If the field makes an angle of 50.0° with the normal to the plane of the loop, determine the magnetic flux through the loop.

Faraday’s Law: Experiments
A current appears only if there is relative motion between the loop and the magnet; the current disappears when the relative motion between them ceases. Faster motion produces a greater current. If moving the magnet’s north pole toward the loop causes, say, clockwise current, then moving the north pole away causes counterclockwise current. Moving the south pole toward or away from the loop also causes currents, but in the reversed directions. An emf is induced in the loop when the number of magnetic field lines that pass through the loop is changing.

The needle deflects momentarily when the switch is closed

Faraday’s Law of Induction
The magnitude of the emf induced in a conducting loop is equal to the rate at which the magnetic flux through that loop changes with time， If a coil consists of N loops with the same area, the total induced emf in the coil is given by In uniform magnetic field, the induced emf can be expressed as

To induce an emf we can change,
the magnitude of B the area enclosed by the loop the angle between B and the normal to the area any combination of the above over time.

Example 1: EMF in a loop A wire loop of radius 0.30m lies so that an external magnetic field of strength +0.30T is perpendicular to the loop. The field changes to -0.20T in 1.5s. (The plus and minus signs here refer to opposite directions through the loop.) Find the magnitude of the average induced emf in the loop during this time.

The loop is always perpendicular to the field, so the normal to the loop is parallel to the field, so cos q = 1. The flux is then Initially the flux is and after the field changes the flux is The magnitude of the average induced emf is:

Example: One way to Induce an emf in a coil
A coil consists of 200 turns of wire having a total resistance of 2.0 . Each turn is a square of side 18 cm, and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly form 0 to 0.50 T in 0.80s, what is the magnitude of the induced emf in the coil while the field is changing. The area of one turn of the coil is (0.18m)2 = m2. The magnetic flux through the coil at t=0 is zero because B=0 at that time. At t=0.80s, the magnetic flux through one turn is : B = BA = (0.50T)(0.0324m2) = T.m2 Therefore, the magnitude of the induced emf is :

Example : An Exponentially Decaying B Field
A loop of wire enclosing an area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magnitude of B varies in time according to the expression B = Bmax e-at, where a is some constant. That is, at t=0 the field is Bmax, and for t>0, the field decreases exponentially. Find the induced emf in the loop as a function of time.

Solution Because B is perpendicular to the plane of the loop, the magnetic flux thought the loop at time t>0 is : B = BAcos0 = ABmaxe-at Because Abmax and a are constants, the induced emf is : This expression indicates that the induced emf decays exponentially in time. Note that the maximum emf occurs at t=0, where max = aABmax.

Motional EMF As the wire moves,
Which sets the charges in motion in the direction of FB and leaves positive charges behind. As they accumulate on the bottom, an electric field is set up inside. In equilibrium,

Motional EMF in a Circuit
If the bar is moved with constant velocity,

Example: Motional emf Induced in a Rotating Bar
A conducting bar of length L rotates with a constant angular speed ω about a pivot at one end. A uniform magnetic field B is directed perpendicular to the plane of rotation, as shown in Figure. Find the motional emf induced between the ends of the bar.

Solution Consider a segment of the bar of length dr having a velocity v. The magnitude of the emf induced in this segment is : Because every segment of the bar is moving perpendicular to B, an emf d of the same form is generated across each. Summing the emfs induced across all segments, which are in series, gives the total emf between the ends of the bar : To integrate this expression, we must note that the linear speed of an element is related to the angular speed  through the relationship v = r. Therefore, because B and  are constants, we find that :

Example : Magnetic Force Acting on a Sliding Bar
The conducting bar illustrated in Figure, of mass m and length , moves on two frictionless parallel rails in the presence of a uniform magnetic field directed into the page. The bar is given an initial velocity vi to the right and is released at t=0. Find the velocity of the bar as a function of time.

Solution The induced current is counterclockwise, and the magnetic force is FB = -I B, where the negative sign denotes that the force is to the left and retards the motion. This is the only horizontal force acting on the bar, and hence Newton’s second law applied to motion in the horizontal direction gives : We know that I=B v/R, and so we can write this expression as :

Integrating this equation using the initial condition that v=vi at t=0, we find that:
where the constant From this result, we see that the velocity can be expressed in the exponential form : This expression indicates that the velocity of the bar decreases exponentially with time under the action of the magnetic retarding force.

Lenz’s Law The polarity of the induced emf is such that it tends to produce a current that creates a magnetic flux to oppose the change in magnetic flux through the area enclosed by the current loop. As the bar is slid to the right, the flux through the loop increases. This induces an emf that will result in an opposing flux. Since the external field is into the screen, the induced field has to be out of the screen. Which means a counterclockwise current

Energy Considerations
Suppose, instead of flowing counterclockwise, the induced current flows clockwise: Then the force will be towards the right which will accelerate the bar to the right which will increase the magnetic flux which will cause more induced current to flow which will increase the force on the bar … and so on All this is inconsistent with the conservation of energy

Moving Magnet and Stationary Coil
Right moving magnet increases flux through the loop. It induces a current that creates it own magnetic field to oppose the flux increase. Left moving magnet decreases flux through the loop. It induces a current that creates it own magnetic field to oppose the flux decrease.

Application of Lenz’s Law
When the switch is closed, the flux goes from zero to a finite value in the direction shown. To counteract this flux, the induced current in the ring has to create a field in the opposite direction. After a few seconds, since there is no change in the flux, no current flows. When the switch is opened again, this time flux decreases, so a current in the opposite direction will be induced to counter act this decrease.

A Loop Moving Through a Magnetic Field

Induced EMF and Electric Fields
Electric Field Inside a Conductor Changing Magnetic Flux EMF This induced electric field is non-conservative and time-varying General Form of Faraday’s Law

Example : Electric Field Induced by a Changing Magnetic Field in a Solenoid
A long solenoid of radius R has n turns of wire per unit length and carries a time-varying current that varies sinusoidally as I = Imax cos t, where Imax is the maximum current and  is the angular frequency of the alternating current source Figure. (a) Determine the magnitude of the induced electric field outside the solenoid, a distance r > R from its long central axis. (b) What is the magnitude of the induced elelctric field inside the solenoid, a distance r from its axis?

Solution for (a) Consider an external point and take the path for our line integral to be a circle of radius r centered on the solenoid as illustrated in Figure (31.18). The magnitude of E is constant on this path and that E is tangent to it. The magnetic flux through the area enclosed by this path is BA = BR2; hence : (1) The magnetic field inside a long solenoid is given by , B = onI. Substitute I = Imax cos t into this equation and then substitute the result into Eq. (1), we find that :

(2) (for r > R)

Solution for (b) For an interior point (r < R), the flux threading an integration loop is given by Br2. Using the same procedure as in part (a), we find that : (3) (for r < R)

Generators and Motors

Example: An AC generator consists of 8 turns of wire, each of area A = m2 , and the total resistance of the wire is 12.0 (. The loop rotates in a T magnetic field at a constant frequency of 60.0 Hz. (A) Find the maximum induced emf. (B) What is the maximum induced current when the output terminals are connected to a low-resistance conductor?

Maxwell’s Equations Gauss’ Law
Gauss’ Law for Magnetism no magnetic monopoles Faraday’s Law Ampère-Maxwell Law Lorentz Force Law

[1] A rectangular coil of 150 loops forms a closed circuit with a resistance of 5 and measures 0.2 m wide by 0.1 m deep, as shown below. The circuit is placed between the poles of an electromagnet which is producing a uniform magnetic field of 40 T. The magnet is switched off, causing the magnetic field to drop to zero in 2 s. (The loops are parallel with the faces of the electromagnet.) a) Compute the average induced potential in the circuit. b) Determine the average current in the circuit. c) Indicate on the drawing the direction in which the induced current flows.

C] The magnetic field from the magnet is up, but the flux is decreasing. So the magnetic field from the current induced in the loop(s) will also be up. Thus the current flow is left-to-right across the front of the loop.

(b) What is the mechanical power delivered by the constant force?
[2] A conducting rod of length l moves on two (frictionless) horizontal rails, as shown to the right. A constant force of magnitude moves the bar at a constant speed of through a magnetic field directed into the page. The resistor has a value (a) What is the current through the resistor R? (b) What is the mechanical power delivered by the constant force?

When the conducting rod moves to the right, this serves to increase the flux as time passes , so any induced current wants to stop this change and decrease the magnetic flux. Therefore, the induced current will act in such a way to oppose the external field (i.e., the field due to the induced current will be opposite to the external field). This must be a counterclockwise current. To find the current, we only need to find the motionally-induced voltage and then apply Ohm’s law. The rod is just a bar of length l moving at velocity v in a magnetic field B. This gives us an voltage ∆V , and Ohm’s law gives us I:

Keep in mind that v is velocity, while ∆V is voltage
Keep in mind that v is velocity, while ∆V is voltage. Of course, the problem is now that we don’t know B. We do know that the external and magnetic forces must balance for the rod to have a constant velocity however. Constant velocity implies zero acceleration, which implies no net force. If this is to be true, the applied force must exactly balance the magnetic force on the rod. For a conductor of length l carrying a current I in a field B, we know how to calculate the magnetic force Plug that into the first equation: You should get I=0.5 A.

What about the power? Conservation of energy tells us that the mechanical power delivered must be the same as the power dissipated in the resistor, or You should get 2 W. Alternatively, you note that power delivered by a force is where θ is the angle between the force and velocity. In this case, θ = 0, so

[4] Figure shows a right view of a bar that can slide without friction
[4] Figure shows a right view of a bar that can slide without friction. The resistor is 6.00 Ω and a 2.50T magnetic field is directed perpendicularly downward, into the paper. Let L= 1.20 m. (a) Calculate the applied force required to move the bar to the right at a constant speed of 2.00 m/s. (b) At what rate is energy delivered to the resistor?

[6] Consider the mass spectrometer
[6] Consider the mass spectrometer. The magnitude of the electric field between the plates of the velocity selector is V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of T. Calculate the radius of the path for a singly charged ion having a mass m =2.18 * kg.

[7] An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT. The angular momentum of the electron about the center of the circle is 4.00 * Js. Determine (a) the radius of the circular path and (b) the speed of the electron.

[8] A proton moves with a velocity of v = (2iˆ - 4jˆ +kˆ ) m/s in a region in which the magnetic field is B = (iˆ +2jˆ - 3kˆ ) T. What is the magnitude of the magnetic force this charge experiences? Give it in unit vector notation,

General Physics II ELECTRICITY AND MAGNETISM

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General Physics II ELECTRICITY AND MAGNETISM

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