Lecture 9 Applications of Gauss’s Law Conductor in electric field.

Lecture 9

Applications ofGauss’s Law

Conductor in electric field

ACT: Crossed planes

Which diagram corresponds to the E-field lines for these two uniformly charged infinite sheets that intersect each other as shown?

+y

+x

+y

+x

+y

+x

A B C

Each sheet produces a uniform electric field.

+x

+y

totalE

E

E

The total E field is uniform in each quadrant.

+y

+x

EXAMPLE: Infinite line of charge

A cable of diameter D = 3 mm and length L = 200 m has a total charge Q = 4 C uniformly distributed along its length. Find the electric field at point P, located at a distance r = 2 cm from it.

r

P

L >> r → infinite cable

r >> D → one-dimensional charge distribution

Linear charge density:

QL

The elegant way (Gauss’s law)

r

P

h

The system has cylindrical symmetry.

The Gaussian surface should be a cylinder of radius r and height h.

r

P

h

capscylinder side

E

caps 0

(caps are parallel to lines)E

cylinder side 2E rh

Linear charge density:

QL

enclosedq h

0

h

r

P

h

0

2h

rhE

Linear charge density:

QL

02E

r

Example: Line and sheet

A. 0

B. 2.05 × 105 N/C

C. 4.10 × 105 N/C

D. 6.15 × 105 N/C

E. 9.25 × 105 N/C

Find the magnitude of the electric field half-way between an infinite line of uniform charge (λ = +10 μC/m) that runs parallel to an infinite sheet of uniform charge (σ = +10 μC/m2). The distance between the line and the sheet is d = 1.0 m.

λ = +10 μC/m

σ = +10 μC/m2

P d/2 d/2

enclosed0 0

0 0

2

2 2

EAA

EA EQ A

A

2r

Electric field produced by an infinite sheet at distance r :

σ

enclosed0 0

0 0

22

2 2

E rLL k

E rL EQ Lr r

Electric field produced by an infinite line at distance r:

L

rλ

λ = +10 μC/m

σ = +10 μC/m2

P d/2 d/2

Esheet Eline

sheet0

line

2

2

E

kE r

r

P line sheet

0 0

0

5

( )2

2

1

2

2.05 10 N/ C

dE E E

d

d

(Answer B)

Charge in a conductor

We know that E = 0 inside a conductor in equilibrium. Therefore, the electric flux though any Gaussian surface inside the conductor is zero.

There is no charge inside the conductor.

The charge enclosed by these surfaces is zero.

The charge in a conductor in equilibrium is always on the surface(s).

Second application of Gauss’s Law:

Finding the charge distribution when E is known.

Example: Conducting shell

A solid non-conducting sphere with a total charge Q = +3 μC is surrounded by a concentric uncharged conducting spherical shell.What is the surface charge density σin on the inner surface of the shell?

σoutσin

Q

σoutσin

Q

Negative charges are attracted towards the inner surface and positive charges are repelled towards the outer surface.

σoutσin

Q

Draw a Gaussian surface inside the conducting shell.

0 ( 0 inside a conductor)E da E

Q +Qin

=0

Qin

=−QRin

=q

enclosed

0

=Q +Q

in

0

Charge on the outer surface:

Net charge on the shell is zero: Q

out+Q

in=0

Q

out=−Q

in=Q

Q-Q

Q

The charge enclosed by the pink surface must be zero.

If the metal shell had a total charge of 3Q :

4Q-Q

Q

If the metal shell had a total charge of 3Q and the central charge was 100Q :

103Q-100Q

100Q

The charge enclosed by the pink surface must be zero.

ACT: Conducting shell

We now remove the uncharged shell. Compare the magnitude of the electric field at point P before and after the shell is removed.

A. Ebefore < Eafter

B. Ebefore = Eafter

C. Ebefore > Eafter

σoutσin

Q

P

σin

Q

σoutσin

Q

P

In both cases, the symmetry is the same. We will use the same Gaussian surface: A sphere that contains point P.

And the enclosed charge is also the same: Q.

Therefore,

2P

0

20 P

4

14

QE r

QE

r

2P4E r So the integral will look just the same:

ACT: Asymmetric conducting shell

This time keep the shell, but move the internal charge off center. Compare the electric field at point P when the charge is centered/off center.

A. Ecentered < Eoff-center

B. Ecentered = Eoff-center

C. Ecentered > Eoff-center Q

P

Q

P

QQ

Charge distribution on inner surface:

Uniform Concentrated near the sphere

But E = 0 inside the

shell in both cases!

Outer surface charge is

uniform (Why wouldn’t it?)

E is the same for both (that of a

sphere with uniform charge

Q)

Flux through pink surface =

0

-Q -Q

Total charge on inner surface is always -Q.

QQ

Total charge on outer surface is always Q.

ACT: Parallel charged planes II

An uncharged metal slab is inserted between the two planes as shown below. Charge densities σL and σR appear on the sides of the slab. Compare the electric field at point P before and after the slab is introduced.

A. Ebefore < Eafter B. Ebefore = Eafter C. Ebefore > Eafter

P

σL σR

σL + σR = 0and distance from the plane does not matter!

σ2 = -2 μC/m2σ1 = +4 μC/m2

P

σL σR

How can we determine σL and σR?

σ1

σ2

The slab surfaces are two new charged planes. There are 4 contributions to the field inside the slab

1

02

2

02

L

02

R

02

Note: σ2 < 0, so that contribution really points in the opposite direction, but it’s easier to leave signs for the end.

P

σL σRσ1

σ2

• The net field inside the conducting slab is zero:

1

02

2

02

L

02

R

02

L R1 2

0 0 0 0

02 2 2 2

E

1 L R 2

L R

0

0

1 L 22 0

R L

2 1

L 2

1 2

R 2