Gauss’s Law

PH 203

Professor Lee Carkner

Lecture 5

Gauss

= q/0 = ∫EdA

Note that: Flux only depends on net q internal to surface

For a uniform surface and uniform q, E is the

same everywhere on surface so ∫EdA = EA

Cylinder E field is always radially outward

We want to find E a distance r away

To solve Gauss’s Law: = q/0 = ∫EdA = EA q is h

Solve for E

a)

Plane

We can again capture the flux with a cylindrical Gaussian surface

Useful for large sheet or point close to sheet

b)

Spherical Shell

Consider a spherical shell of charge of radius r and total charge q

c)

d)

Surface within Sphere What if we have total charge q,

uniformly distributed with a radius R?

e)

What if surface is inside R?

If we apply r3/R3 to the point charge formula we get,

E = (q/40R3)r

Conductors and Charge

The charges in the conductor are free to move and so will react to each other

Like charges will want to get as far away from each other as possible

No charge inside conductor

Charge Distribution How does charge distribute itself over a surface?

e.g., a sphere

No component parallel to surface, or else the charges would move

Excess charge there may spark into the air

Conductors and External

The positive charges will go to the surface “upfield” and the negative will go to the surface “downfield”

The field inside the conductor is zero The charges in the conductor cancel out the external

field A conductor shields the region inside of it

Conducting Ring

Charges

Pushed

To

Surface

No E Field

InsideField Lines

Perpendicular

to Surface

Faraday Cage

If we make the conductor hollow we can sit inside it an be unaffected by external fields

Your car is a Faraday cage and is thus a good place to be in a thunderstorm

Next Time

Read 24.1-24.6 Problems: Ch 23, P: 24, 36, 45, Ch 24, P:

2, 4

A uniform electric field of magnitude 1 N/C is pointing in the positive y direction. If the cube has sides of 1 meter, what is the flux through sides A, B, C?

A) 1, 0, 1B) 0, 0, 1C) 1, 0, 0D) 0, 0, 0E) 1, 1, 1

A

B

C

Consider three Gaussian surfaces. Surface 1 encloses a charge of +q, surface 2 encloses a charge of –q and surface 3 encloses both charges. Rank the 3 surfaces according to the flux, greatest first.

A) 1, 2, 3

B) 1, 3, 2

C) 2, 1, 3

D) 2, 3, 1

E) 3, 2, 1

+q -q

1 2

3

Rank the following Gaussian surfaces by the amount of flux that passes through them, greatest first (q is at the center of each).

A) 1, 2, 3

B) 1, 3, 2

C) 2, 1, 3

D) 3, 2, 1

E) All tie

q q

1

2

q

3

Rank the following Gaussian surfaces by the strength of the field at the surface at the point direction below q (where the numbers are).

A) 1, 2, 3

B) 1, 3, 2

C) 2, 1, 3

D) 3, 2, 1

E) All tie

q q

1

2

q

3