Two conductors in proximity form a “capacitor”: they have a capacity to hold a charge Q (+Q on one and -Q on the other) with a voltage difference V. C=Q/V

Two conductors in proximity form a “capacitor”: they have a capacity to hold a charge Q (+Q on one and -Q on the other)

with a voltage difference V.

C=Q/V

Note: Capacitance C is italicized. Coulomb C is not. (Sorry!)

Capacitance is a property of the geometry of the conductors (and depends on whether there is vacuum or material

between them, as we will see later.)

This is because of linearity (superposition.) If I double the source charges, I will double the field, and double the

potential.

If I put ±2 microcoulombs of charge on the sphere and plate, there is a potential difference of 2 V.

What is the capacitance?

A] 0

B] Cannot determine, since not symmetric

C] 1 microfarad = 10-6 C/V

D] 106 farads

If I put ±2 microcoulombs of charge on the sphere and plate, there is a potential difference of 2 V.

If I put ±4 microcoulombs of charge on the sphere and plate, what will be the potential difference?

A] 0

B] 2 V

C] 4 V

D] 8 V

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 d

By superposition, the field in the dielectric is the sum of the original capacitor field, plus the field from the polarization surface charge.

The total field is therefore

A] bigger than the original capacitor field

B] smaller than the original capacitor field

C] the same as the original capacitor field

Answer: smaller than the original field.

We can write that the new field is the original field, divided by K, the “dielectric constant”

• What is the new potential across the capacitor?

A] same as the old potential, without the dielectric

B] bigger than the old potential, by a factor of K

C] smaller than the old potential, by a factor of K

If I put a charge ±Q on a parallel plate capacitor and increase the plate separation, what happens to V?

A] it increases

B] it decreases

C] it stays the same


The field is the same, but the distance is larger. So V increases.

What happens to the potential energy U?

A] it increases

B] it decreases



The field is the same, but the distance is larger. So V increases.

What happens to the potential energy U?

A] it increases

The energy density u is the same, but there is more volume.Where did this energy come from??

If I put a charge ±Q on a parallel plate capacitor and insert a dielectric, what happens to V?

A] it increases

B] it decreases



The field in the dielectric is reduced by a factor of K. So the potential goes down.

What happens to the potential energy?

A] it goes up by a factor of K

B] it goes up by a factor of K2

C] it goes down by a factor of K

D] it goes down by a factor of K2

E] it stays the same.


What happens to the potential energy?It may be counterintuitive, but the potential energy goes down by a

factor of K (not K2).

U=Q2/(2C); the capacitance goes up by a factor of K.

Note: the energy density for a given E field in a dielectric is

The field goes down by a factor of K, but epsilon adds a factor of K.

€

u = 12 εE

2

If the potential energy goes down, where does the energy go?

• The field pulls the dielectric into the capacitor, giving it kinetic energy. (Very small, here.)

• Application: Optical tweezers

Batteries don’t store charge. They store energy.

A chemical battery works because electrons like to leave some materials to go to others.

The change in energy when the electron goes “downhill” becomes available as electric POTENTIAL between

the battery terminals

Lead Acid (Car) Battery

The rxn on the right will not go in the direction indicated unless the electrolyte sol’n potential is closer than 1.685 V to the + electrode potential.

i.e. the + electrode can be no more than 1.685 V higher in potential than the electrolyte sol’n.

Pb+2 is in the form of solid lead sulfate on the electrodes. When “discharged”, both electrodes turn into lead sulfate.


The rxn on the left will not go in the direction indicated unless the electrolyte potential is closer than 0.356 V to the - electrode potential.

i.e. the - electrode can be no more than 0.356 V lower in potential than the electrolyte soln.


Overall, then, the reactions will STOP when the potential between the +/- electrodes is 2.041V.

Only a tiny tiny tiny amount of charge needs to build up on the electrodes for the reaction to stop.How much?

But if you connect the electrodes with a resistive wire, the reaction will start to go as the potential drops a hair below 2.041V.


What happens if you force the potential difference to be higher than 2.041 V?

The reactions run backwards!Lead sulfate turns into lead oxide and lead (metallic).

This is charging the battery.


Note that the rxn doesn’t make a big “reservoir” of electrons.

The battery doesn’t die because it runs out of stored electrons.

It doesn’t store electrons. It “pumps” electrons “on demand”, i.e. when the potential falls below 2.041 V.

The two electrodes of an ideal V volt battery are shown. The field lines for the electric field between the electrodes are shown when nothing is

attached to the electrodes. The electrodes have a separation = d.

A resistive wire of length L is then attached to the battery.What is the electric field in the wire, when steady state is reached?

A] 0

B] it varies, but averages to V/d

C] it varies, but averages to V/L

D] it is V/L everywhere in the wire

E] no way to determine

The two electrodes of an ideal V volt battery are shown. The field lines for the electric field between the electrodes are shown when nothing is attached to the electrodes. The electrodes have a separation = d.

A resistive wire of length L is then instantaneously attached to the battery.What is the electric field in the wire, immediately after attaching it?

A] 0

B] it varies, but averages to V/d

C] it varies, but averages to V/L

D] it is V/L everywhere in the wire

E] no way to determine

Where on this wire will negative charges “build up” (a tiny amount)

A

B

Assume wires have zero resistance

Assume wires have zero resistance

No E field in electrostatics.

A] same

B] 3R

C] 9R

D] R/3

E] R/9

On left, both are equally bright. Bulb A on the right is brighter, As V^2/R is bigger.

If the R’s are all the same, which arrangement has the lowest Ref ? B

Which has the highest Ref ? A

Which has a lower Ref:

C or d? D

If the 2 cm long resistor is 3 ohms, what is the resistance of the 1 cm long resistor (with half the radius)? (Both are of the same material.)

A] 1 ohm

B] 1.5 ohms

C] 2 ohms

D] 3 ohms

E] 6 ohms

What is the current through the battery?

A] 0 amps

B] 0.5 amps

C] 1 amp

D] 2 amps

E] 3 amps

What is the power delivered by the battery? In watts

A] 0

B] 1

C] 2

D] 3

E] 4

What is the magnitude of the E field the physically larger resistor (3 ohms)? (in V/m)

A] 0

B] 100

C] 200

D] 300

E] 400

What is the magnitude of the E field the physically smaller resistor (6 ohms)? (in V/m)

A] 0

B] 100

C] 200

D] 300

E] 400

What is the current through the battery? In amps

A] 0

B] 1/9

C] 1/6

D] 2/9

E] 2/3

What is the E field in the 3 ohm resistor? (in V/m)

A] 0

B] 16

C] 33 1/3

D] 78

E] 200

Are R1 and R2 in parallel, in series, or neither?

A] parallelB] seriesC] neither

Which junction formula is correct?

Which loop formula is correct?

At steady state, what is current I1?

At steady state, if the voltage at a = 0, what is the voltage at b?

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Two conductors in proximity form a “capacitor”: they have a capacity to hold a charge Q (+Q on one and -Q on the other) with a voltage difference V. C=Q/V