With insulators, you can often use Gauss’ law in combination with the principle of superposition.

(PoS is that the field from two sources is the vector sum of the fields from each source.)

Two examples:1) Two infinite planes, one with charge density +, one with charge density - .

2) Subtractive superposition: a sphere with a spherical (off center!) void

What is the direction of the electric field at the point shown?

A] NE

B] SE

C] NW

D] SW

E]it varies, depending on distances a & b

What is the magnitude of the electric field at the point shown?

A]

B]

C]

D]

E] 0

€

2ε0

€

ε0

€

2σ

ε0

€

2σ

2ε0

Use extreme caution when applying Gauss’ law, when conductors are present.

It is important not only that the “initial state” have sufficient symmetry, but also that the final state maintain the symmetry.

For example, consider an infinite plane with initial charge density , and then add a point charge above it.

If the plane is insulating, you can solve this problem. If the plane is conducting, you can’t solve this problem!

Electric Fields & Conductors (in statics…)

No field in the conductor (or else charges would move)

So, by Gauss’ law, no excess charge inside conductor (all excess charge is on surfaces)

Field lines must contact the conductor perpendicularly

The surface charge density is directly proportional to the field just outside the conductor (by Gauss’ law) E = ε0

Let’s draw some field lines for

a conducting plane with -Q total charge, and an external point charge of +Q

& conducting shells with charges inside.

There still must be no field in the conductor. So the inner surface is still -Q. Thus, the outer surface will have +9Q, so that the total is 8Q (=+9Q - 1Q).

Given the peanut shaped geometry and the fact that the charge +Q is off-center, is the charge density on the inner surface uniform (the same everywhere on that surface)?

a] yes

b] no

The negative charge on the inner surface will be concentrated close to the positive charge. (The E field next to the surface is stronger there !)

If there is +9Q of charge on the outer surface, will the charge density on the outer surface be uniform?

a] yes

b] no

There is no electric field in the conductor. So the outside of the conductor cannot “know” that there is an off-center charge in the middle!

So the charge density on the outer surface is uniform.

a] yes

A metal box with no net charge is placed in an initially uniform E field, as shown.

What is the total charge on the inner surface ? Assume this surface has area A.

A] 0

B]

C]

D]

E] cannot determine

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ε0EA

€

−ε0EA

€

2ε0EA

No charge on the inner surface. What is the total charge on the outer surface?

A] 0

B]

C]

D]

E] cannot determine

€

ε0EA

€

−ε0EA

€

2ε0EA

No charge on the inner surface. Since the total charge is 0, there can be no net charge on the outer surface.

What is the charge density on the left outer face of the box?Assume the external field is uniform, as shown.

A] 0

B]

C]

D]

E] cannot determine

€

ε0E

€

−ε0E

€

2ε0E

What is the electric flux of the field

through the rectangle in the xz plane between x=[0,L1], z=[0,L2]

A] 0

B] L12L2

C] larger than answer B

D] smaller than answer B

What is the electric flux of the field

through the half pipe shown:

A] 0

B] L12L2

C] larger than answer B

D] smaller than answer B

What is the flux of the electric field through the surface shown?

A] 0

B]

C]

D]

E]

€

λL

ε0

€

πλL

ε0

€

λL

8ε0

€

2λL

ε0

A point charge +Q is a distance d above an insulating sheet with charge density .

What is the field at point P?

Insulating sheet: superposition of fields gives answer B.

Suppose, instead, that a conducting sheet with charge density = is brought from far away (far down, in the picture) to a distance d away from the charge +Q, then what is the field at P?

With a conducting sheet, the charge +Q will cause the charges to redistribute. Cannot determine! (Need Physics 400 level)

Suppose, instead, that a conducting sheet with charge density = 0 is brought from far away (far down, in the picture) to a distance d away from the charge +Q, then what is the field at P?

In which case does the electric potential energy increase?

A B

Or C: both cases D: neither case

In which case does the electric potential energy increase?

A B

Or C: both cases D: neither case

Work done by electric force (source: fixed charges)on a test charge

depends only on endpoints, not on path.

(You can see this easily for a single fixed charge… it holds in general because of superposition.)

Electric forces are “conservative” - We can define a potential energy.

When a + charge moves “down the field”, the electric force does work on it, increasing its kinetic energy (or putting energy elsewhere).

When a + charge moves “up the field”, it either loses kinetic energy, or some other force must push it up.

The electrical potential energy of a system of charges is the work necessary to assemble the charges from “infinity”.

(For point charges, we take U=0 at infinity.)

This will include all pairs of interactions.

Two equal + charges are initially stationary and separated by r0.

If they are allowed to fly apart (to infinity), what will be the kinetic energy of each?

€

1

4πε0

q

r0

€

1

4πε0

q2

r0€

1

4πε0

q2

r02

€

1

8πε0

q2

r0

A] B]

C] D]

Three equal + charges are initially stationary and at the vertices of an equilateral triangle with side r0.

If they are allowed to fly apart (to infinity), what will be the kinetic energy of each?

€

1

4πε0

q

r0

€

1

4πε0

q2

r0€

1

4πε0

q2

r02

€

1

8πε0

q2

r0

A] B]

C] D]

Just as we can define electric field as the force felt by a test charge

We define “potential” as potential energy of a test charge.

Just as a conservative force is: (minus) the derivative of the potential energy

The electric field is(minus) the derivative of the potential.

€

W = −ΔU = − dU∫ = Fdx∫€

Vpc =1

4πε0

qpc

r

€

dV =1

4πε0

dq

r

Eqiupotential surfaces

Equipotential surfaces are perpendicular to field lines

In 2D pics, equipotentials look like lines, but they are surfaces.

Note: just because V=0 doesn’t mean E=0!

A function can be zero but have a non-zero derivative.

Also: it’s time to think in 3D. The derivative can be taken w.r.t x, y, or z.

€

−dV =r E • d

v l

€

−dV = Exdx + Eydy + E zdz

€

Ex = −∂V

∂xThis means: hold y, z constant, so dy=dz=0

Note that E is a vector, but V is a scalar.

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vE = −

v ∇V

3 ways to calculate E fields

A) Direct sum of sources, using Coulomb + Calculus (+ Components!)B) Gauss’ Law & SymmetryC) The negative of the derivative of the electric potential (if given)

2 Ways to calculate the electric potentialA) The negative of the integral of the E field (if given)B) Sum of the sources, using Calculus if necessary

Note: by sum of sources, I mean use the result from integrating the Coulomb field for a point source, V =

€

kq

r

Last time, we found the potential from a ring of charge.

Here’s another example of integrating over sources to find V:

A line of charge.

Let’s find V by integrating E for a line of charge.

A perfectly insulating plane has charge density+2 C/m2.

What is the magnitude of the E field a distance x above the plane (in terms of x andε?

A] ε

B] 1/ε

C] 2/ε

D] 2εx

F] 2/(xε

Where is the electrical potential higher?

A] farther from the plane of positive charge

B] closer to the plane of positive charge

C] it’s the same everywhere, since the plane is infinite

The field is 1.13 x 1011 N/C, regardless of x. (Gauss)

What is the magnitude of the difference in potential between a point in the plane and a point 10 m above the plane?

A] 1.13 x 1011 V

B] 1.13 x 1010 V

C] 1.13 x 109 V

D] 0 V

E] potential difference does not exist in this problem since V ≠ 0 at infinity

Which graph could be the potential from an infinite plane of positive charge density, where x = distance from plane?

E (none)