Presentation on theme: "Electric Charge and Electric Field. Static Electricity; Electric Charge and Its Conservation Electric Charge in the Atom Insulators and Conductors Induced."— Presentation transcript:
Electric Charge and Electric Field
Static Electricity; Electric Charge and Its Conservation Electric Charge in the Atom Insulators and Conductors Induced Charge; the Electroscope Coulomb’s Law The Electric Field Electric Field Calculations for Continuous Charge Distributions
Field Lines Electric Fields and Conductors Motion of a Charged Particle in an Electric Field Electric Dipoles Electric Forces in Molecular Biology: DNA Photocopy Machines and Computer Printers Use Electrostatics
Charge comes in two types, positive and negative; like charges repel and opposite charges attract. Electric Charge and Its Conservation
Electric charge is conserved – the arithmetic sum of the total charge cannot change in any interaction. Electric Charge and Its Conservation
Atom: Nucleus (small, massive, positive charge) Electron cloud (large, very low density, negative charge) Electric Charge in the Atom
Polar molecule: neutral overall, but charge not evenly distributed Electric Charge in the Atom
Conductor: Charge flows freely Metals Insulator: Almost no charge flows Most other materials Some materials are semiconductors. Insulators and Conductors
Electric Charge I 1) one is positive, the other is negative 2) both are positive 3) both are negative 4) both are positive or both are negative Two charged balls are repelling each other as they hang from the ceiling. What can you say about their charges?
1) have opposite charges 2) have the same charge 3) all have the same charge 4) one ball must be neutral (no charge) From the picture, what can you conclude about the charges? Electric Charge II
Conductors I 1) positive 2) negative 3) neutral 4) positive or neutral 5) negative or neutral A metal ball hangs from the ceiling by an insulating thread. The ball is attracted to a positive-charged rod held near the ball. The charge of the ball must be:
Two neutral conductors are connected by a wire and a charged rod is brought near, but does not touch. The wire is taken away, and then the charged rod is removed. What are the charges on the conductors? ConcepTest 21.2bConductors II ConcepTest 21.2b Conductors II 1)00 2)+– 3)–+ 4)++ 5)– – 0 0 ? ?
Experiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them. Coulomb’s Law
Coulomb’s law: This equation gives the magnitude of the force between two charges. Coulomb’s Law
Unit of charge: coulomb, C. The proportionality constant in Coulomb’s law is then: k = 8.99 x 10 9 N·m 2 /C 2. Charges produced by rubbing are typically around a microcoulomb: 1 μC = C. Coulomb’s Law
Charge on the electron: e = x C. Electric charge is quantized in units of the electron charge. Coulomb’s Law
The proportionality constant k can also be written in terms of ε 0, the permittivity of free space: Coulomb’s Law
The force is along the line connecting the charges, and is attractive if the charges are opposite, and repulsive if they are the same. Coulomb’s Law
Q Q F 1 = 3 N F 2 = ? 1) 1.0 N 2) 1.5 N 3) 2.0 N 4) 3.0 N 5) 6.0 N What is the magnitude of the force F 2 ? Coulomb’s Law I
In its vector form, the Coulomb force is: Coulomb’s Law
Which charge exerts the greater force? Two positive point charges, Q 1 = 50 μC and Q 2 = 1 μC, are separated by a distance. Which is larger in magnitude, the force that Q 1 exerts on Q 2 or the force that Q 2 exerts on Q 1 ?
Coulomb’s Law Three charges in a line. Three charged particles are arranged in a line, as shown. Calculate the net electrostatic force on particle 3 (the -4.0 μC on the right) due to the other two charges.
3R3R +Q+Q – – 4Q Two balls with charges +Q and –4Q are fixed at a separation distance of 3R. Is it possible to place another charged ball Q 0 anywhere on the line such that the net force on Q 0 will be zero? Electric Force III Q 0 is positive 1) yes, but only if Q 0 is positive Q 0 is negative 2) yes, but only if Q 0 is negative Q 0 3) yes, independent of the sign (or value) of Q 0 4) no, the net force can never be zero
Coulomb’s Law Electric force using vector components. Calculate the net electrostatic force on charge Q 3 shown in the figure due to the charges Q 1 and Q 2.
Which of the arrows best represents the direction of the net force on charge +Q due to the other two charges? +2Q +4Q +Q+Q d d Forces in 2D
Coulomb’s Law Make the force on Q 3 zero. In the figure, where could you place a fourth charge, Q 4 = -50 μC, so that the net force on Q 3 would be zero?
The electric field is defined as the force on a small charge, divided by the magnitude of the charge: The Electric Field
An electric field surrounds every charge.
For a point charge: The Electric Field
Force on a point charge in an electric field: The Electric Field
A photocopy machine works by arranging positive charges (in the pattern to be copied) on the surface of a drum, then gently sprinkling negatively charged dry toner (ink) particles onto the drum. The toner particles temporarily stick to the pattern on the drum and are later transferred to paper and “melted” to produce the copy. Suppose each toner particle has a mass of 9.0 x kg and carries an average of 20 extra electrons to provide an electric charge. Assuming that the electric force on a toner particle must exceed twice its weight in order to ensure sufficient attraction, compute the required electric field strength near the surface of the drum.
The Electric Field Electric field of a single point charge. Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = -3.0 x C.
What is the electric field at the center of the square? C 5) E = 0 Superposition I
C Superposition II What is the electric field at the center of the square? 5) E = 0
What is the direction of the electric field at the position of the X ? Q+Q -Q-Q +Q+Q 5 Superposition III
The Electric Field E at a point between two charges. Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?
The Electric Field E above two point charges. Calculate the total electric field (a) at point A and (b) at point B in the figure due to both charges, Q 1 and Q 2.
Problem solving in electrostatics: electric forces and electric fields 1. Draw a diagram; show all charges, with signs, and electric fields and forces with directions. 2. Calculate forces using Coulomb’s law. 3. Add forces vectorially to get result. 4. Check your answer! The Electric Field
Continuous Charge Distributions A continuous distribution of charge may be treated as a succession of infinitesimal (point) charges. The total field is then the integral of the infinitesimal fields due to each bit of charge: Remember that the electric field is a vector; you will need a separate integral for each component.
Continuous Charge Distributions A ring of charge. A thin, ring-shaped object of radius a holds a total charge +Q distributed uniformly around it. Determine the electric field at a point P on its axis, a distance x from the center. Let λ be the charge per unit length (C/m).
Solution: Because P is on the axis, the transverse components of E must add to zero, by symmetry.
Continuous Charge Distributions Charge at the center of a ring. Imagine a small positive charge placed at the center of a nonconducting ring carrying a uniformly distributed negative charge. Is the positive charge in equilibrium if it is displaced slightly from the center along the axis of the ring, and if so is it stable? What if the small charge is negative? Neglect gravity, as it is much smaller than the electrostatic forces.
Continuous Charge Distributions Long line of charge. Determine the magnitude of the electric field at any point P a distance x from a very long line (a wire, say) of uniformly distributed charge. Assume x is much smaller than the length of the wire, and let λ be the charge per unit length (C/m).
Solution: The components of E parallel to the wire must add to zero by symmetry.
Continuous Charge Distributions Uniformly charged disk. Charge is distributed uniformly over a thin circular disk of radius R. The charge per unit area (C/m 2 ) is σ. Calculate the electric field at a point P on the axis of the disk, a distance z above its center.
Solution: The disk is a set of concentric rings and, for a ring with a radius r, we know its contribution to the electric field r z θ dr
Continuous Charge Distributions In the previous example, if we are very close to the disk (that is, if z << R ), the electric field is: This is the field due to an infinite plane of charge. Infinite plane
Continuous Charge Distributions Two parallel plates. Determine the electric field between two large parallel plates or sheets, which are very thin and are separated by a distance d which is small compared to their height and width. One plate carries a uniform surface charge density σ and the other carries a uniform surface charge density -σ as shown (the plates extend upward and downward beyond the part shown).
The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge. Field Lines
A proton and an electron are held apart a distance of 1 m and then let go. Where would they meet? 1) in the middle 2) closer to the electron’s side 3) closer to the proton’s side p e Proton and Electron III
1) charges are equal and positive 2) charges are equal and negative 3) charges are equal and opposite 4) charges are equal, but sign is undetermined 5) charges cannot be equal Q2Q2 Q1Q1 x y E Two charges are fixed along the x axis. They produce an electric field E directed along the negative y axis at the indicated point. Which of the following is true? Find the Charges
The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge. The electric field is stronger where the field lines are closer together. Field Lines
Electric dipole: two equal charges, opposite in sign: Field Lines
The electric field between two closely spaced, oppositely charged parallel plates is constant. Field Lines uniform field
Q In a uniform electric field in empty space, a 4 C charge is placed and it feels an electric force of 12 N. If this charge is removed and a 6 C charge is placed at that point instead, what force will it feel? 1) 12 N 2) 8 N 3) 24 N 4) no force 5) 18 N Uniform Electric Field
Electric Field Lines I What are the signs of the charges whose electric fields are shown at right? 1) 2) 3) 4) 5) no way to tell
Electric Field Lines II Which of the charges has the greater magnitude? 1) 2) 3) both the same
Summary of field lines: 1.Field lines indicate the direction of the field; the field is tangent to the line. 2.The magnitude of the field is proportional to the density of the lines. 3.Field lines start on positive charges and end on negative charges; the number is proportional to the magnitude of the charge. Field Lines
The static electric field inside a conductor is zero – if it were not, the charges would move. The net charge on a conductor resides on its outer surface. Conductors
The electric field is perpendicular to the surface of a conductor – again, if it were not, charges would move. Conductors
Shielding, and safety in a storm. A neutral hollow metal box is placed between two parallel charged plates as shown. What is the field like inside the box?
Motion in an Electric Field The force on an object of charge q in an electric field is given by: = q Therefore, if we know the mass and charge of a particle, we can describe its subsequent motion in an electric field.
Motion in an Electric Field Electron accelerated by electric field. An electron (mass m = 9.11 x kg) is accelerated in the uniform field ( E = 2.0 x 10 4 N/C) between two parallel charged plates. The separation of the plates is 1.5 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. (a) With what speed does it leave the hole? (b) Show that the gravitational force can be ignored. Assume the hole is so small that it does not affect the uniform field between the plates.
Motion in an Electric Field Electron moving perpendicular to. Suppose an electron traveling with speed v 0 = 1.0 x 10 7 m/s enters a uniform electric field, which is at right angles to v 0 as shown. Describe its motion by giving the equation of its path while in the electric field. Ignore gravity.
Electric Dipoles An electric dipole consists of two charges Q, equal in magnitude and opposite in sign, separated by a distance. The dipole moment, p = Q, points from the negative to the positive charge.
Electric Dipoles An electric dipole in a uniform electric field will experience no net force, but it will, in general, experience a torque:
Electric Dipoles The electric field created by a dipole is the sum of the fields created by the two charges; far from the dipole, the field shows a 1/r 3 dependence:
Electric Dipoles Dipole in a field. The dipole moment of a water molecule is 6.1 x C·m. A water molecule is placed in a uniform electric field with magnitude 2.0 x 10 5 N/C. (a) What is the magnitude of the maximum torque that the field can exert on the molecule? (b) What is the potential energy when the torque is at its maximum? (c) In what position will the potential energy take on its greatest value? Why is this different than the position where the torque is maximum?
Two kinds of electric charge – positive and negative. Charge is conserved. Charge on electron: e = x C. Conductors: electrons free to move. Insulators: nonconductors. Summary
Charge is quantized in units of e. Objects can be charged by conduction or induction. Coulomb’s law: Electric field is force per unit charge: Summary
Electric field of a point charge: Electric field can be represented by electric field lines. Static electric field inside conductor is zero; surface field is perpendicular to surface. Summary
Electric Flux Gauss’s Law Applications of Gauss’s Law Experimental Basis of Gauss’s and Coulomb’s Laws
Electric flux: Electric flux through an area is proportional to the total number of field lines crossing the area. Electric Flux
Electric flux. Calculate the electric flux through the rectangle shown. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle θ is 30°.
Flux through a closed surface: Electric Flux
The net number of field lines through the surface is proportional to the charge enclosed, and also to the flux, giving Gauss’s law: This can be used to find the electric field in situations with a high degree of symmetry. Gauss’s Law
For a point charge, Therefore, Solving for E gives the result we expect from Coulomb’s law:
Gauss’s Law Using Coulomb’s law to evaluate the integral of the field of a point charge over the surface of a sphere surrounding the charge gives: Looking at the arbitrarily shaped surface A 2, we see that the same flux passes through it as passes through A 1. Therefore, this result should be valid for any closed surface.
Gauss’s Law Finally, if a gaussian surface encloses several point charges, the superposition principle shows that: Therefore, Gauss’s law is valid for any charge distribution. Note, however, that it only refers to the field due to charges within the gaussian surface – charges outside the surface will also create fields.
Gauss’s Law Flux from Gauss’s law. Consider the two gaussian surfaces, A 1 and A 2, as shown. The only charge present is the charge Q at the center of surface A 1. What is the net flux through each surface, A 1 and A 2 ?
Applications of Gauss’s Law Spherical conductor. A thin spherical shell of radius r 0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere?
Applications of Gauss’s Law Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r 0. Determine the electric field (a) outside the sphere ( r > r 0 ) and (b) inside the sphere ( r < r 0 ).
Applications of Gauss’s Law Nonuniformly charged solid sphere. Suppose the charge density of a solid sphere is given by ρ E = α r 2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r 0. (b) Find the electric field as a function of r inside the sphere.
Applications of Gauss’s Law Long uniform line of charge. A very long straight wire possesses a uniform positive charge per unit length, λ. Calculate the electric field at points near (but outside) the wire, far from the ends.
Solution: Since the wire is essentially infinite, it has cylindrical symmetry and we expect the field to be perpendicular to the wire everywhere.
Applications of Gauss’s Law Infinite plane of charge. Charge is distributed uniformly, with a surface charge density σ (σ = charge per unit area = dQ/dA ) over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane.
Solution: We expect E to be perpendicular to the plane.
Applications of Gauss’s Law Electric field near any conducting surface. Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by E = σ / ε 0 where σ is the surface charge density on the conductor’s surface at that point.
Applications of Gauss’s Law The difference between the electric field outside a conducting plane of charge and outside a nonconducting plane of charge can be thought of in two ways: 1. The field inside the conductor is zero, so the flux is all through one end of the cylinder. 2. The nonconducting plane has a total charge density σ, whereas the conducting plane has a charge density σ on each side, effectively giving it twice the charge density.
Applications of Gauss’s Law Conductor with charge inside a cavity. Suppose a conductor carries a net charge + Q and contains a cavity, inside of which resides a point charge + q. What can you say about the charges on the inner and outer surfaces of the conductor?
Applications of Gauss’s Law Procedure for Gauss’s law problems: 1. Identify the symmetry, and choose a gaussian surface that takes advantage of it (with surfaces along surfaces of constant field). 2. Draw the surface. 3. Use the symmetry to find the direction of E. 4. Evaluate the flux by integrating. 5. Calculate the enclosed charge. 6. Solve for the field.
Electric flux: Gauss’s law: Gauss’s law can be used to calculate the field in situations with a high degree of symmetry. Gauss’s law applies in all situations, and therefore is more general than Coulomb’s law. Summary