copyright © 2009 pearson education, inc. chapter 22 gauss’s law

Copyright © 2009 Pearson Education, Inc.

Chapter 22Gauss’s Law


Flux through a closed surface:

22-1 Electric Flux


Think in terms of flux lines:


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.

22-2 Gauss’s Law


22-2 Gauss’s Law

For a point charge,

Therefore,

Solving for E gives the result we expect from Coulomb’s law:

Spherical


22-2 Gauss’s LawUsing 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 A2, we see that the same flux passes through it as passes through A1. Therefore, this result should be valid for any closed surface.


22-2 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.


22-2 Gauss’s Law

Conceptual Example 22-2: Flux from Gauss’s law.

Consider the two gaussian surfaces, A1 and A2, as shown. The only charge present is the charge Q at the center of surface A1. What is the net flux through each surface, A1 and A2?


22-3 Applications of Gauss’s LawExample 22-3: Spherical conductor.

A thin spherical shell of radius r0 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? Spherical


22-3 Applications of Gauss’s Law

Example 22-4: Solid sphere of charge.

An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0).

Spherical/radial



Example 22-5: Nonuniformly charged solid sphere.

Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere.



Example 22-6: 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.

Linear,Cylindrical



Example 22-7: 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.

Planar & cylindrical



Example 22-8: 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/ε0

where σ is the surface charge density on the conductor’s surface at that point.



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.



Conceptual Example 22-9: 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?



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.

��E


22-4 Experimental Basis of Gauss’s and Coulomb’s Laws

In the experiment shown, Gauss’s law predicts that the charge on the ball flows onto the surface of the cylinder when they are in contact. This can be tested by measuring the charge on the ball after it is removed – it should be zero.


• 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 of Chapter 22


Chapter 23Electric Potential


• Electric Potential Energy and Potential Difference

• Relation between Electric Potential and Electric Field

• Electric Potential Due to Point Charges

• Potential Due to Any Charge Distribution

• Equipotential Surfaces

• Electric Dipole Potential

Units of Chapter 23


• E Determined from V

• Electrostatic Potential Energy; the Electron Volt

Units of Chapter 23


The electrostatic force is conservative – potential energy can be defined.

Change in electric potential energy is negative of work done by electric force:

23-1 Electrostatic Potential Energy and Potential Difference

E

electronelectron

protonproton

E

ElectronElectron

ProtonProton++

--

1) proton

2) electron

3) both feel the same acceleration

4) neither – there is no acceleration

5) they feel the same magnitude acceleration but opposite direction

A A protonproton and an and an electronelectron are in are in a constant electric field created a constant electric field created by oppositely charged plates. by oppositely charged plates. You release the You release the protonproton from the from the positivepositive side and the side and the electronelectron from the from the negativenegative side. Which side. Which has the larger acceleration?has the larger acceleration?

ConcepTest 23.1bConcepTest 23.1b Electric Potential Energy II Electric Potential Energy II

E

electronelectron

protonproton

E

ElectronElectron

ProtonProton++

--

1) proton

2) electron

3) both feel the same acceleration

4) neither – there is no acceleration

5) they feel the same magnitude acceleration but opposite direction

Since F = maF = ma and the electron is much less electron is much less

massivemassive than the proton, the electron electron

experiences the larger accelerationexperiences the larger acceleration.

A A protonproton and an and an electronelectron are in are in a constant electric field created a constant electric field created by oppositely charged plates. by oppositely charged plates. You release the You release the protonproton from the from the positivepositive side and the side and the electronelectron from the from the negativenegative side. Which side. Which has the larger acceleration?has the larger acceleration?

ConcepTest 23.1bConcepTest 23.1b Electric Potential Energy II Electric Potential Energy II


Electric potential is defined as potential energy per unit charge:

Unit of electric potential: the volt (V):

1 V = 1 J/C.



Only changes in potential can be measured, allowing free assignment of V = 0:



Analogy between gravitational and electrical potential energy:




Conceptual Example 23-1: A negative charge.

Suppose a negative charge, such as an electron, is placed near the negative plate at point b, as shown here. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change?



Electrical sources such as batteries and generators supply a constant potential difference. Here are some typical potential differences, both natural and manufactured:


23-2 Relation between Electric Potential and Electric Field

The general relationship between a conservative force and potential energy:

Substituting the potential difference and the electric field:



The simplest case is a uniform field:



Example 23-3: Electric field obtained from voltage.

Two parallel plates are charged to produce a potential difference of 50 V. If the separation between the plates is 0.050 m, calculate the magnitude of the electric field in the space between the plates.



Example 23-4: Charged conducting sphere.

Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for (a) r > r0, (b) r = r0, (c) r < r0. The total charge on the sphere is Q.



The previous example gives the electric potential as a function of distance from the surface of a charged conducting sphere, which is plotted here, and compared with the electric field:


23-3 Electric Potential Due to Point Charges

To find the electric potential due to a point charge, we integrate the field along a field line:



Setting the potential to zero at r = ∞ gives the general form of the potential due to a point charge:



Example: Work required to bring two positive charges close together.

What minimum work must be done by an external force to bring a proton q = 1.60×10-19 C from a great distance away (take r = ∞) to a point 1.00×10-15 m from another proton?



Example: Work required to bring two positive charges close together.

What minimum work must be done by an external force to bring a proton q = 1.60×10-19 C from a great distance away (take r = ∞) to a point 1.00×10-15 m from another proton?

W = ke2/r = (9×109) (1.6×10-19)2/ (1×10-15) = 2.3×10-13 J = 2.3×10-13 J


1) V > 0

2) V = 0

3) V < 0

AA BB

What is the electric What is the electric

potential at point B?potential at point B?

ConcepTest 23.3 Electric Potential


Since Q2 and Q1 are equidistant

from point B, and since they have

equal and opposite charges, the

total potential is zerozero.

1) V > 0

2) V = 0

3) V < 0

AA BB

What is the electric What is the electric

potential at point B?potential at point B?

ConcepTest 23.3 Electric Potential

Follow-up:Follow-up: What is the potential What is the potential at the origin of the at the origin of the x yx y axes? axes?



Example 23-7: Potential above two charges.

Calculate the electric potential (a) at point A in the figure due to the two charges shown.


23-4 Potential Due to Any Charge Distribution

The potential due to an arbitrary charge distribution can be expressed as a sum or integral (if the distribution is continuous):

or



Example 23-8: Potential due to a ring of charge.

A thin circular ring of radius R has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring a distance x from its center.



Example 23-9: Potential due to a charged disk.

A thin flat disk, of radius R0, has a uniformly distributed charge Q. Determine the potential at a point P on the axis of the disk, a distance x from its center.


Which of these configurations gives V = 0 at all points on the x axis?

4) all of the above 5) none of the above

1)

x

+2C

-2C

+1C

-1C2)

x

+2C

-1C

+1C

-2C3)

x

+2C

-1C

-2C

+1C

ConcepTest 23.5 Equipotential Surfaces


Only in case (1), where opposite charges lie

directly across the x axis from each other, do

the potentials from the two charges above the

x axis cancel the ones below the x axis.

Which of these configurations gives V = 0 at all points on the x axis?

4) all of the above 5) none of the above

1)

x

+2C

-2C

+1C

-1C2)

x

+2C

-1C

+1C

-2C3)

x

+2C

-1C

-2C

+1C

ConcepTest 23.5 Equipotential Surfaces


An equipotential is a line or surface over which the potential is constant.

Electric field lines are perpendicular to equipotentials.

The surface of a conductor is an equipotential

(E// =0 → ∂V/∂s// = 0)

23-5 Equipotential Surfaces


23-5 Equipotential SurfacesEquipotential surfaces are always perpendicular to field lines; they are always closed surfaces (unlike field lines, which begin and end on charges).


23-5 Equipotential Surfaces

A gravitational analogy to equipotential surfaces is the topographical map – the lines connect points of equal gravitational potential (altitude).


The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. For distances large compared to the charge separation:

23-6 Electric Dipole Potential


23-7 E Determined from V

If we know the field, we can determine the potential by integrating. Inverting this process, if we know the potential, we can find the field by differentiating:

This is a vector differential equation; here it is in component form:

��

E


23-7 E Determined from V

Example 23-11: E for ring and disk.

Use electric potential to determine the electric field at point P on the axis of (a) a circular ring of charge and (b) a uniformly charged disk.

��

E��E


Homework Assignment # 3

Chapter 22 – 10, 22, 34

Chapter 23 – 14, 30, 36

Tentative HW # 4:

Chapter 24 – 6, 16, 28, 46, 60, 82

copyright © 2009 pearson education, inc. chapter 22 gauss’s law

Documents