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

€

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 density2 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

Suppose the field were 9 x 109 N/C.

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

A] 9 x 109 V

B] 9 x 1010 V

C] 9 x 1011 V

D] 0 V

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

Now consider a wire and a plane…

You can just add the potential differences from each source, but you need to be careful with signs.

Rank the magnitudes (smallest to largest) of the electric field at point P in the three arrangements shown.

A] all are the same

B] I, II, III

C] III, II, I

D] II, I, III

Rank the electric potentials at point P (smallest to largest). A

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

E (none)

Which graph could be the potential from an infinite line of positive charge density, where x = distance from line? E (should be logarithmic, decreasing)

E (none)

Which graph could be the potential from a positive point charge, where x = distance from charge? B

E (none)

What is the magnitude of the E field at Q?

A] 1 V/m

B] 2 V/m

C] 4 V/m

D] 6 V/m

E] none of these

What is the magnitude of the potential at Q?

Take V=0 at infinity.

A] 1 V

B] 2 V

C] 4 V

D] 6 V

E] none of these

Use a pencil & paper. The potential at Q (half way between the identical charges) is 30 V. (Take V=0 at infinity.)

What is the potential at P, one-quarter of the separation away from one charge?

A] 15 V

B] 30 V

C] 40 V

D] 60 V

E] cannot determine w/o more info

Use a pencil & paper. The potential at Q (half way between the identical charges) is 30 V. (Take V=0 at infinity.)

VP = 40 V.

If these charges are held fixed, and a 1 C charge is released from point P, what is its kinetic energy at point Q?

A] zero J B] 1 J

C] 10 J D] 40 J

E] none of these

Use pencil & paper. A point charge Q sits a distance L above an infinite plane of charge density Q/L2.

Where is the E field zero?

Use pencil & paper. A point charge Q sits a distance L above an infinite plane of charge density Q/L2.

Where is the potential zero?

Use pencil & paper. What is the change in potential moving from point b, L/2 below the plane, to point a, L/2 above the plane, in line with the charge Q?

Use pencil & paper. What is the change in potential moving between a point c, in the positively charged plane, to point a at L/2 above the plane, in line with the charge Q?

More practice finding E from V

If V = -4x + 4y2 (x,y in meters, V in volts) , what is the E field at the origin?

A] 0

B] 4 V/m in the +y direction

C] 4 V/m in the -y direction

D] 4 V/m in the +x direction

E] 4 V/m in the -x direction

If V = -4x + 4y2 (x,y in meters, V in volts) , where is the E field = 0?

A] at the origin

B] at x=1, y=1

C] at x=1, y=-1

D] at both B and C

E] nowhere.

Equipotential Lines = Contours of constant V

E fieldpoints downhill

Downhill is always perpendicular to level

Conductors at rest are equipotential

Two infinite parallel sheets carry charge densities ± What is the electric field at point 1?

Two infinite parallel sheets carry charge densities ± What is the electric field at point 2?

The potential difference between the plates is dε