a

b

c

•Choose either or

•And E constant over surface

• is just the area of the Gaussian surface over which we are integrating.

•Gauss’ Law

•This equation can now be solved for E (at the surface) if we know qenclosed (or for qenclosed if we know E).

Gauss’ Law

E dS E dS EdS

0E dS E dS

Given an infinite sheet of charge as shown in the figure. You need to use Gauss' Law to calculate the electric field near the sheet of charge. Which of the Gaussian surfaces are best suited for this purpose?

• a cylinder with its axis along the plane?

– No, E not parallel or perpendicular to the cylinder

• a cylinder with its axis perpendicular to the plane?

– Yes

• a cube?

– Yes

• a sphere?

– No

Gaussian Surfaces

Infinite sheet of charge

• Symmetry:

direction of E = x-axis

• Therefore, CHOOSE Gaussian surface to be a cylinder whose axis is aligned with the x-axis.

Ax

E E

Conclusion: An infinite plane sheet of charge creates a CONSTANT electric field .

• Apply Gauss' Law:

• The charge enclosed = A

Therefore, Gauss’ Law

• On the barrel,• On the ends,

Two Infinite Sheets(into the screen)

• Field outside must be zero. Two ways to see:

– Superposition

– Gaussian surface encloses zero charge

E=0 E=0

E

+

+

++

+

+

+

+

+

+

-

-

--

-

--

-

-

-+

+

---

A

A

0

• Field inside is NOT zero:

– Superposition

– Gaussian surface encloses non-zero charge

Question 1

The figure shows a Gaussian surface enclosing charges 2q and –q. The net flux through the surface is:

a) q/0 b) 2q/0 c)-q/0 d)zero

2q

-q

Question 1

1. a

2. b

3. c

4. d

Question 1

The figure shows a Gaussian surface enclosing charges 2q and –q. The net flux through the surface is:

a) q/0 b) 2q/0 c)-q/0 d)zero

2q

-q

•The net charge enclosed is +q. Gauss’ Law says the flux through the surface is q/0

•q of the lines from the positive charge go to the negative charge

•q go to infinity and thus pass through the surface

• How to do practically all of the homework problems

• Gauss’ Law is ALWAYS VALID!

Gauss’ Law: Help for the Problems

• What Can You Do With This?If you have symmetry (a) spherical, (b) cylindrical, or (c) planar AND:

• If you know the charge (RHS), you can calculate the electric field (LHS)• If you know the field (LHS, usually because E=0 inside conductor), you can calculate the charge (RHS).

Gauss’ Law: Help for the Problems

q = ALL charge inside radius r

• Spherical Symmetry: Gaussian surface = Sphere of radius r

q = ALL charge inside cylinder =A

• Planar Symmetry: Gaussian surface = Cylinder of area A

q = ALL charge inside radius r, length L

• Cylindrical symmetry: Gaussian surface = cylinder of radius r

Example 1: spheres• A solid conducting sphere is concentric

with a thin conducting shell, as shown

• The inner sphere carries a charge Q1, and the spherical shell carries a charge Q2,

such that Q2 = -3Q1.

• How is the charge distributed on the sphere?

• How is the charge distributed on the spherical shell?

• What is the electric field at r < R1?

Between R1 and R2? At r > R2?

• What happens when you connect the two spheres with a wire? (What are the

charges?)

R1

R2

Q1

Q2=-3Q1

A

B

C

D

• How is the charge distributed on the sphere?A

* The electric field inside a conductor is zero.

(A) By Gauss’s Law, there can be no net

charge inside the conductor, and the charge

must reside on the outside surface of the

sphere

+

+

+

+ +

+

++

R1

R2

Q1

Q2=-3Q1

• How is the charge distributed on the spherical shell?

B

* The electric field inside the conducting shell is zero.

(B) There can be no net charge inside the conductor,

therefore the inner surface of the shell must carry a

net charge of -Q1, to cancel the inner conductor

charge.

The outer surface must carry the charge +Q1 + Q2, so

that the net charge on the shell equals Q2.

The charges are distributed uniformly over the inner

and outer surfaces of the shell, hence

and22

1inner

R4

Q

22

12

2

12outer

R4

Q2

R4

QQ

R1

R2

Q1

Q2=-3Q1

(C) r < R1:

Inside the conducting sphere

* The electric field inside a conductor is zero.

.0E

(C) Between R1 and R2 : R1 < r < R2

Charge enclosed = Q1

1

20

1ˆ

4

QE r

r

(C) r > R2

Charge enclosed = Q1 + Q2

1 2 1

2 20 0

1 1 2ˆ ˆ

4 4

Q Q QE r r

r r

• What is the Electric Field at r < R1? Between R1 and R2? At r > R2?

C

R1

R2

Q1

Q2=-3Q1

D • What happens when you connect the two spheres with a wire? (What are the

charges?)

After electrostatic equilibrium is reached, there is no charge on the inner sphere, and none on the inner surface of the shell. The charge Q1 + Q2 on the outer surface remains.

--

---

-

- -

---

--

- - - - --

-

--

Also, for r < R2 .0E

and for r > R21

20

21ˆ

4

QE r

r

R1

R2

Q1

Q2=-3Q1

Are Gauss’ and Coulomb’s Laws Correct?

The table shows the results of such experiments looking for a deviation from an inverse-square law:

Are Gauss’ and Coulomb’s Laws Correct?

•One problem with the above experiments is that they have all been done at short range, 1 meter or so.

•Other experiments, more sensitive to cosmic-scale distances, have been done, testing whether Coulomb’s law has the form:

•No evidence for a nonzero μ has been found.

Example 2: Cylinders

An infinite line of charge passes directly through the middle of a hollow, charged, CONDUCTING infinite cylindrical shell of radius R. We will focus on a segment of the cylindrical shell of length h. The line charge has a linear charge density , and the cylindrical shell has a net surface charge density of total.

outer

R

inner

h

total

outer

h

R

total

inner

•How is the charge distributed on the cylindrical shell?

•What is inner?•What is outer?

•What is the electric field at r<R?

•What is the electric field for r>R?

A

B

C

outer

h

R

total

inner

•The electric field inside the cylindrical shell is zero. •Choose as our Gaussian surface a cylinder, which lies inside the cylindrical shell.•Then the net charge enclosed is zero. •Therefore, there will be a surface charge density on the inside wall of the cylinder to balance out the charge along the line.

A1 What is inner?

outer

htotal

inner

R

•The total charge on the enclosed portion (of length h) of the line charge is h

•Therefore, the charge on the inner surface of the conducting cylindrical shell is

•The total charge is evenly distributed along the inside surface of the cylinder. •Therefore, the inner surface charge density inner is just Qinner divided by total area of the cylinder: •Notice that the result is independent of h.

A1 What is inner?

2 2inner

h

Rh R

innerQ h

outer

h

R

total

inner

•We know that the net charge density on the cylinder is total.

• The charge densities on the inner and outer surfaces of the

cylindrical shell have to add up to total.

•Therefore,

outer =total – inner = total + /(2R).

What is outer? A2

h

Gaussian surface

What is the Electric Field at r<R?

•Whenever we are dealing with electric fields created by symmetric charged surfaces, we must always first chose an appropriate Gaussian surface.

•In this case, for r <R, the appropriate surface surrounding the line

charge is a cylinder of radius r.

r

R

h

B

h

Gaussian surface

What is the Electric Field at r<R?

Using Gauss’ Law, the E field is given by

qenclosed is the charge on the enclosed line charge h, (2rh) is the area of the barrel of the Gaussian surface.

The Field at radius r is:

r

R

h

B

0 0

2enclosedr r

q hAE rhE

•Chose a Gaussian surface as indicated above. •Charge enclosed in our Gaussian surface•Net charge enclosed on the line: h•Net charge on the cylindrical shell: = 2Rh total

•Therefore, net charge enclosed is •The surface area of the Gaussian surface is 2rh•Gauss’ Law:

h

R

r

Gaussian surface

total

What is the Electric field for r>R?C

•Solve for Er to find

0

22 total

r r

Rh hAE rhE

2 totalRh h

Summary

• Practice Gauss’ Law problems• Remember SYMMETRY, CONSTANT FIELD• Choose appropriate Gaussian surface

• Next week Electric Potential. Read Chapter 24• Try Chapter 23 problems 25, 33, 47, 51, 56 • Next week: Quiz on Thursday and Friday will be on

chapters 21,22,23

0

enclosedqEA