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**Electric Forces and Electric Fields**

Chapter 19 Electric Forces and Electric Fields

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**Electricity and Magnetism, Some History**

Many applications Macroscopic and microscopic Chinese Documents suggests that magnetism was observed as early as 2000 BC Greeks Electrical and magnetic phenomena as early as 700 BC Experiments with amber and magnetite

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**Electricity and Magnetism, Some History, 2**

1600 William Gilbert showed electrification effects were not confined to just amber The electrification effects were a general phenomena 1785 Charles Coulomb confirmed inverse square law form for electric forces

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**Electricity and Magnetism, Some History, 3**

1820 Hans Oersted found a compass needle deflected when near a wire carrying an electric current 1831 Michael Faraday and Joseph Henry showed that when a wire is moved near a magnet, an electric current is produced in the wire

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**Electricity and Magnetism, Some History, 4**

1873 James Clerk Maxwell used observations and other experimental facts as a basis for formulating the laws of electromagnetism Unified electricity and magnetism 1888 Heinrich Hertz verified Maxwell’s predictions He produced electromagnetic waves

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**Electric Charges There are two kinds of electric charges**

Called positive and negative Negative charges are the type possessed by electrons Positive charges are the type possessed by protons Charges of the same sign repel one another and charges with opposite signs attract one another

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**Electric Charges, 2 The rubber rod is negatively charged**

The glass rod is positively charged The two rods will attract

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**Electric Charges, 3 The rubber rod is negatively charged**

The second rubber rod is also negatively charged The two rods will repel

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**More About Electric Charges**

The net charge in an isolated system is always conserved For example, charge is not created in the process of rubbing two objects together The electrification is due to a transfer of electrons from one object to another

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**Quantization of Electric Charges**

The electric charge, q, is said to be quantized q is the standard symbol used for charge as a variable Electric charge exists as discrete packets q = N e N is an integer e is the fundamental unit of charge |e| = 1.6 x C Electron: q = -e Proton: q = +e

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**Conservation and Quantization of Electric Charges, Example**

A glass rod is rubbed with silk Electrons are transferred from the glass to the silk Each electron adds a negative charge to the silk An equal positive charge is left on the rod The charges on the two objects are ±e, or ±2e, …

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**Three objects are brought close to one another, two at a time**

Three objects are brought close to one another, two at a time. When objects A and B are brought together, they repel. When objects B and C are brought together, they also repel. Which of the following statements are true? Objects A and C possess charges of the same sign. Objects A and C possess charges of opposite sign. All three objects possess charges of the same sign. One of the objects is neutral. We need to perform additional experiments to determine the signs of the charges.

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Conductors Electrical conductors are materials in which some of the electrons move relatively freely Free electrons are not bound to the atoms These electrons can move relatively freely through the material Examples of good conductors include copper, aluminum and silver When a good conductor is charged in a small region, the charge readily distributes itself over the entire surface of the material

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Insulators Electrical insulators are materials in which electric charges do not move freely Examples of good insulators include glass, rubber and wood When a good insulator is charged in a small region, the charge is unable to move to other regions of the material

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Semiconductors The electrical properties of semiconductors are somewhere between those of insulators and conductors Examples of semiconductor materials include silicon and germanium The electrical properties of semiconductors can be changed over many orders of magnitude by adding controlled amounts of foreign atoms to the materials

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Charging by Induction Charging by induction requires no contact with the object inducing the charge Assume we start with a neutral metallic sphere The sphere has the same number of positive and negative charges

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Charging by Induction, 2 A negatively charged rubber rod is placed near the sphere It does not touch the sphere The electrons in the neutral sphere are redistributed The migration of electrons leaves the side near the rod with an effective positive charge

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**Charging by Induction, 3 The sphere is grounded**

Grounded means the conductor is connected to an infinite reservoir for electrons, such as the Earth Some electrons can leave the sphere through the ground wire

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**Charging by Induction, 4 The ground wire is removed**

There will now be more positive charges in the sphere The positive charge has been induced in the sphere

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**Charging by Induction, 5 The rod is removed**

The rod has lost none of its charge during this process The electrons remaining on the sphere redistribute themselves There is still a net positive charge on the sphere

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**Charge Rearrangement in Insulators**

A process similar to induction can take place in insulators The charges within the molecules of the material are rearranged The effect is called polarization

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Charles Coulomb 1736 – 1806 Major contributions in the fields of electrostatics and magnetism Also investigated Strengths of materials Structural mechanics Ergonomics How people and animals can best do work

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Coulomb’s Law Charles Coulomb measured the magnitudes of electric forces between two small charged spheres He found the force depended on the charges and the distance between them Torsion balance

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Coulomb’s Law, 2 The electrical force between two stationary charged particles is given by Coulomb’s Law The force is inversely proportional to the square of the separation r between the particles and directed along the line joining them The force is proportional to the product of the charges, q1 and q2, on the two particles

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Point Charge The term point charge refers to a particle of zero size that carries an electric charge The electrical behavior of electrons and protons is well described by modeling them as point charges

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**Coulomb’s Law, Equation**

Mathematically, The SI unit of charge is the Coulomb, C ke is called the Coulomb Constant ke = x 109 N.m2/C2 = 1/(4peo) eo is the permittivity free space eo = x C2 / N.m2

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**Coulomb's Law, Notes Remember the charges need to be in Coulombs**

e is the smallest unit of charge Except quarks e = 1.6 x C So 1 C needs 6.24 x 1018 electrons or protons Typical charges can be in the µC range Remember that force is a vector quantity

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**Vector Nature of Electric Forces**

In vector form, is a unit vector directed from q1 to q2 The like charges produce a repulsive force between them

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**Vector Nature of Electrical Forces, 2**

Electrical forces obey Newton’s Third Law The force on q1 is equal in magnitude and opposite in direction to the force on q2 With like signs for the charges, the product q1q2 is positive and the force is repulsive With opposite signs for the charges, the product q1q2 is negative and the force is attractive

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**Vector Nature of Electrical Forces, 3**

Two point charges are separated by a distance r The unlike charges produce an attractive force between them With unlike signs for the charges, the product q1q2 is negative and the force is attractive

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**A Final Note About Directions**

The sign of the product of q1q2 gives the relative direction of the force between q1 and q2 The absolute direction is determined by the actual location of the charges

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Hydrogen Atom Example The electrical force between the electron and proton is found from Coulomb’s Law Fe = keq1q2 / r2 = 8.2 x 10-8 N This can be compared to the gravitational force between the electron and the proton Fg = Gmemp / r2 = 3.6 x N

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**The Superposition Principle**

The resultant force on any one particle equals the vector sum of the individual forces due to all the other individual particles Remember to add the forces as vectors The resultant force on q1 is the vector sum of all the forces exerted on it by other charges:

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Problem 19.5. Three point charges are located at the corners of an equilateral triangle as shown in Figure P19.5. Calculate the resultant electric force on the 7.00-μC charge.

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**Zero Resultant Force, Superposition Example**

Where is the resultant force equal to zero? The magnitudes of the individual forces will be equal Directions will be opposite Will result in a quadratic Choose the root that gives the forces in opposite directions

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**Electric Field – Test Particle**

The electric field is defined in terms of a test particle, qo By convention, the test particle is always a positive electric charge The test particle is used to detect the existence of the field It is also used to evaluate the strength of the field The test charge is assumed to be small enough not to disturb the charge distribution responsible for the field

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**Electric Field – Definition**

An electric field is said to exist in the region of space around a charged object This charged object is the source particle When another charged object, the test charge, enters this electric field, an electric force acts on it Analogy to gravity: gravity field of the Earth exists around it independent of the objects; however if an object of test mass m is placed in the Earth’s gravity field, force of gravity equal to mg will act on it.

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**Electric Field – Definition, cont**

The electric field is defined as the electric force on the test charge per unit charge The electric field vector, , at a point in space is defined as the electric force, , acting on a positive test charge, qo placed at that point divided by the test charge:

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**Electric Field, Notes is the field produced by some charge or charge**

distribution, separate from the test charge The existence of an electric field is a property of the source charge The presence of the test charge is not necessary for the field to exist The test charge serves as a detector of the field and its strength Test charge does not disturb the charge distribution responsible for the electric field

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**Relationship Between F and E**

This is valid for a point charge only One of zero size For larger objects, the field may vary over the size of the object If q is positive, and are in the same direction If q is negative, and are in opposite directions

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**Electric Field Notes, Final**

The direction of is that of the force on a positive test charge The SI units of are N/C We can also say that an electric field exists at a point if a test charge at that point experiences an electric force

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**Electric Field, Vector Form**

Remember Coulomb’s Law, between the source and test charges, can be expressed as Then, the electric field will be

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**More About Electric Field Direction**

a) q is positive, the force is directed away from q b) The direction of the field is also away from the positive source charge c) q is negative, the force is directed toward q d) The field is also toward the negative source charge

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**changes in a way that cannot be determined.**

A test charge of +3 μC is at a point P where an external electric field is directed to the right and has a magnitude of 4 × 106 N/C. If the test charge is replaced with another charge of −3 μC, the external electric field at P is unaffected reverses direction changes in a way that cannot be determined. 10

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**Superposition with Electric Fields**

At any point P, the total electric field due to a group of source charges equals the vector sum of electric fields at that point due to all the particles Superposition principle is applied in the same manner as for addition of electric forces

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Problem Two point charges are located on the x axis. The first is a charge +Q at x = –a. The second is an unknown charge located at x = +3a. The net electric field these charges produce at the origin has a magnitude of 2keQ/a2. What are the two possible values of the unknown charge?

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**Electric Dipole Example**

Find the electric field due to q1, Find the electric field due to q2, Remember, the fields add as vectors The direction of the individual fields is the direction of the force on a positive test charge Electric dipole

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**Electric Field – Continuous Charge Distribution**

The distances between charges in a group of charges may be much smaller than the distance between the group and a point of interest In this situation, the system of charges can be modeled as continuous The system of closely spaced charges is equivalent to a total charge that is continuously distributed along some line, over some surface, or throughout some volume

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**Electric Field – Continuous Charge Distribution, cont**

Procedure: Divide the charge distribution into small elements, each of which contains Dq Calculate the electric field due to one of these elements at point P Evaluate the total field by summing the contributions of all the charge elements

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**Electric Field – Continuous Charge Distribution, equations**

For the individual charge elements Because the charge distribution is continuous

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Charge Densities Volume charge density – when a charge is distributed evenly throughout a volume r = Q / V Surface charge density – when a charge is distributed evenly over a surface area s = Q / A Linear charge density – when a charge is distributed along a line l = Q / l

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**Example 19.4. The electric field due to a charged rod**

A rod of length l has a uniform linear density and a total charge Q. Calculate the electric field at a point P along the axis of the rod, a distance a from one end.

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**Problem Solving Strategy**

Conceptualize Imagine the type of electric field that would be created by the charges or charge distribution Categorize Analyzing a group of individual charges or a continuous charge distribution? Think about symmetry

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**Problem Solving Hints, cont**

Analyze Group of individual charges: use the superposition principle The resultant field is the vector sum of the individual fields Continuous charge distributions: the vector sums for evaluating the total electric field at some point must be replaced with vector integrals Divide the charge distribution into infinitesimal pieces, calculate the vector sum by integrating over the entire charge distribution

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**Problem Solving Hints, final**

Analyze, cont Symmetry: take advantage of any symmetry in the system Finalize Check to see if your field is consistent with the mental representation and reflects any symmetry Imagine varying parameters to see if the result changes in a reasonable way

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**Electric Field Lines Field lines give us a means**

of representing the electric field pictorially The electric field vector is tangent to the electric field line at each point The line has a direction that is the same as that of the electric field vector The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region

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**Electric Field Lines, General**

The density of lines through surface A is greater than through surface B The magnitude of the electric field is greater on surface A than B The lines at different locations point in different directions This indicates the field is non-uniform

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**Electric Field Lines, Positive Point Charge**

The field lines radiate outward in all directions In three dimensions, the distribution is spherical The lines are directed away from the source charge A positive test charge would be repelled away from the positive source charge

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**Electric Field Lines, Negative Point Charge**

The field lines radiate inward in all directions The lines are directed toward the source charge A positive test charge would be attracted toward the negative source charge

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**Electric Field Lines – Rules for Drawing**

The lines must begin on a positive charge and terminate on a negative charge In the case of an excess of one type of charge, some lines will begin or end infinitely far away The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge Field lines cannot intersect

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**Motion of Charged Particles**

When a charged particle is placed in an electric field, it experiences an electrical force If this is the only force on the particle, it must be the net force The net force will cause the particle to accelerate according to Newton’s Second Law

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**Motion of Particles, cont**

If E is uniform, then a is constant If the particle has a positive charge, its acceleration is in the direction of the field If the particle has a negative charge, its acceleration is in the direction opposite the electric field Since the acceleration is constant, the kinematic equations can be used

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**Electron in a Uniform Field, Example**

The electron is projected horizontally into a uniform electric field The electron undergoes a downward acceleration The charge is negative, so the acceleration is opposite the field Its motion is parabolic while between the plates

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Cathode Ray Tube (CRT) A CRT is commonly used to obtain a visual display of electronic information in oscilloscopes, radar systems, televisions, etc The CRT is a vacuum tube in which a beam of electrons is accelerated and deflected under the influence of electric or magnetic fields

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CRT, cont The electrons are deflected in various directions by two sets of plates The placing of charge on the plates creates the electric field between the plates and allows the beam to be steered

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**Problem 19.54. A small, 2.00-g plastic ball is suspended by a**

20.0-cm-long string in a uniform electric field as shown in the figure. If the ball is in equilibrium when the string makes a 15.0° angle with the vertical, what is the net charge on the ball?

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Electric Flux Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field FE = E A

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**Electric Flux, General Area**

The electric flux is proportional to the number of electric field lines penetrating some surface The field lines may make some angle q with the perpendicular to the surface Then FE = E A cos q

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**Electric Flux, Interpreting the Equation**

The flux is a maximum when the surface is perpendicular to the field The flux is zero when the surface is parallel to the field If the field varies over the surface, F = E A cos q is valid for only a small element of the area

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Electric Flux, General In the more general case, look at a small area element In general, this becomes

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Electric Flux, final The surface integral means the integral must be evaluated over the surface in question In general, the value of the flux will depend both on the field pattern and on the surface The units of electric flux will be N.m2/C2

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**Electric Flux, Closed Surface**

Assume a closed surface The vectors point in different directions At each point, they are perpendicular to the surface By convention, they point outward

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**Flux Through Closed Surface, cont**

At (1), the field lines are crossing the surface from the inside to the outside; q <90o, F is positive At (2), the field lines graze the surface; q =90o, F = 0 At (3), the field lines are crossing the surface from the outside to the inside;180o > q >90o, F is negative

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**Flux Through Closed Surface, final**

The net flux through the surface is proportional to the net number of lines leaving the surface This net number of lines is the number of lines leaving the volume surrounding the surface minus the number entering the volume If En is the component of E perpendicular to the surface, then

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**Gauss’ Law, Introduction**

Gauss’ Law is an expression of the general relationship between the net electric flux through a closed surface and the charge enclosed by the surface The closed surface is often called a Gaussian surface Gauss’ Law is of fundamental importance in the study of electric fields

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Gauss’ Law – General A positive point charge, q, is located at the center of a sphere of radius r The magnitude of the electric field everywhere on the surface of the sphere is E = ke q / r2

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**Gauss’ Law – General, cont.**

The field lines are directed radially outward and are perpendicular to the surface at every point This will be the net flux through the Gaussian surface, the sphere of radius r We know E = kq/r2 and Asphere = 4pr2, so Does not depend on r!

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**Gauss’ Law – General, notes**

The net flux through any closed surface surrounding a point charge, q, is given by q/eo and is independent of the shape of that surface The net electric flux through a closed surface that surrounds no charge is zero Since the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as

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**Gauss’ Law – Final Gauss’ Law states**

qin is the net charge inside the surface represents the total electric field at any point on the surface The total electric field may have contributions from charges both inside and outside of the surface

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For a Gaussian surface through which the net flux is zero, the following four statements could be true. Which of the statements must be true? No charges are inside the surface. The net charge inside the surface is zero. The electric field is zero everywhere on the surface. The number of electric field lines entering the surface equals the number leaving the surface. 10

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**Consider the charge distribution shown in Active Figure 19. 31**

Consider the charge distribution shown in Active Figure (i) What are the charges contributing to the total electric flux through surface S ’? q1 only q4 only q2 and q3 all four charges none of the charges 10

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**q1 only q4 only q2 and q3 all four charges none of the charges 10**

Consider the charge distribution shown in Active Figure (ii) What are the charges contributing to the total electric field at a chosen point on the surface S ’? q1 only q4 only q2 and q3 all four charges none of the charges 10

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Applying Gauss’ Law Gauss’ Law is valid for the electric field of any system of charges or continuous distribution of charge. Although Gauss’ Law can, in theory, be solved to find E for any charge configuration, in practice it is limited to symmetric situations Particularly spherical, cylindrical, or plane symmetry Remember, the Gaussian surface is a surface you choose, it does not have to coincide with a real surface

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**Conditions for a Gaussian Surface**

Try to choose a surface that satisfies one or more of these conditions: The value of the electric field can be argued from symmetry to be constant over the surface The dot product can be expressed as a simple algebraic produce E dA because and are parallel The dot product is 0 because and are perpendicular The field can be argued to be zero everywhere over the surface

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Problem A cylindrical shell of radius 7.00 cm and length 240 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.

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**Example 19.9: The Electric Field Due to a Point Charge**

Choose a sphere as the Gaussian surface E is parallel to dA at each point on the surface

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**Example 19.10: Field Due to a Spherically Symmetric Charge Distribution**

Select a sphere as the Gaussian surface For r> a

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**Spherically Symmetric, cont**

Select a sphere as the Gaussian surface, r < a qin < Q qin = r (4/3pr3)

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**Spherically Symmetric Distribution, final**

Inside the sphere, E varies linearly with r E ® 0 as r ® 0 The field outside the sphere is equivalent to that of a point charge located at the center of the sphere

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**Example 19.11: Field at a Distance from a Line of Charge**

Select a cylindrical charge distribution The cylinder has a radius of r and a length of l E is constant in magnitude and perpendicular to the surface at every point on the curved part of the surface

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**Field Due to a Line of Charge, cont**

The end view confirms the field is perpendicular to the curved surface The field through the ends of the cylinder is 0 since the field is parallel to these surfaces

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**Field Due to a Line of Charge, final**

Use Gauss’ Law to find the field

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**Example 19.12: Field Due to a Plane of Charge**

must be perpendicular to the plane and must have the same magnitude at all points equidistance from the plane Choose a small cylinder whose axis is perpendicular to the plane for the Gaussian surface

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**Field Due to a Plane of Charge, cont**

is parallel to the curved surface and there is no contribution to the surface area from this curved part of the cylinder The flux through each end of the cylinder is EA and so the total flux is 2EA

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**Field Due to a Plane of Charge, final**

The total charge in the surface is sA Applying Gauss’ Law Note, this does not depend on r Therefore, the field is uniform everywhere

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