# Capacitor and dielectric

## Homework Statement

A ##12 \mu F## capacitor ##C_{1}## consists of plates ##A## and ##B##. An ##18 \mu F## capacitor ##C_{2}## consists of plates ##C## and ##D##.

(a) Each are charged up by a ##10 V## battery such that plates ##A## and ##C## are positively charged. They are disconnected from the battery and connected into a series circuit with plate ##A## connected to plate ##D## and plate ##B## connected to plate ##C##, and the system allowed to come to equilibrium. What charge is now on plate ##D##?

(b) Both capacitors ##C_{1}## and ##C_{2}## have square plates of side ##x_{1}## cm and plate separation ##d## cm. The difference is that ##C_1## has only air between its plates, but ##C_2## also has a square plate of dielectric of side ##x_1##, relative permittivity ##\epsilon_{r}## and thickness ##t##. In terms of ##\epsilon_{r}##, what fraction ##(t/d)## of the gap is occupied by the dielectric?

(c) Capacitor ##C_2## is again charged up by a battery of voltage ##V##, and remains connected to it. What force is required to extract the plate of dielectric from the gap? Give answer in terms of ##V## and ##x_1##.

## The Attempt at a Solution

(a) Charge on capacitor ##C_1## is ##Q=CV=120 \mu C##, and charge on capacitor ##C_2## is ##Q=CV=180 \mu C##

Now the arrangement of the capacitors in the series circuit is as follows:

Therefore, after redistribution of charges among plates ##D## and ##A##, an excess negative charge of ##60 \mu C## will remain on the wire connecting plates ##D## and ##A##. Therefore, the charge on plate ##D## is ##-30 \mu C##.

Am I correct so far?

stevendaryl
Staff Emeritus

## Homework Statement

A ##12 \mu F## capacitor ##C_{1}## consists of plates ##A## and ##B##. An ##18 \mu F## capacitor ##C_{2}## consists of plates ##C## and ##D##.

(a) Each are charged up by a ##10 V## battery such that plates ##A## and ##C## are positively charged. They are disconnected from the battery and connected into a series circuit with plate ##A## connected to plate ##D## and plate ##B## connected to plate ##C##, and the system allowed to come to equilibrium. What charge is now on plate ##D##?

(b) Both capacitors ##C_{1}## and ##C_{2}## have square plates of side ##x_{1}## cm and plate separation ##d## cm. The difference is that ##C_1## has only air between its plates, but ##C_2## also has a square plate of dielectric of side ##x_1##, relative permittivity ##\epsilon_{r}## and thickness ##t##. In terms of ##\epsilon_{r}##, what fraction ##(t/d)## of the gap is occupied by the dielectric?

(c) Capacitor ##C_2## is again charged up by a battery of voltage ##V##, and remains connected to it. What force is required to extract the plate of dielectric from the gap? Give answer in terms of ##V## and ##x_1##.

## The Attempt at a Solution

(a) Charge on capacitor ##C_1## is ##Q=CV=120 \mu C##, and charge on capacitor ##C_2## is ##Q=CV=180 \mu C##

Now the arrangement of the capacitors in the series circuit is as follows:

Therefore, after redistribution of charges among plates ##D## and ##A##, an excess negative charge of ##60 \mu C## will remain on the wire connecting plates ##D## and ##A##. Therefore, the charge on plate ##D## is ##-30 \mu C##.

Am I correct so far?

I haven't worked out the answer, but how do you know that the negative charge splits evenly between plates $A$ and $D$?

Oh wait! The voltage across the two capacitors must be the same in magnitude, so the charge of ##-60 \mu C## must divide between plates ##A## and ##D## in the ratio of their capacitances.

##C_{1}:C_{2} = 2:3##, so charge on plate ##D = -36 \mu C##, and charge on plate ##A = -24 \mu C##.

Am I correct so far?

stevendaryl
Staff Emeritus
Oh wait! The voltage across the two capacitors must be the same in magnitude, so the charge of ##-60 \mu C## must divide between plates ##A## and ##D## in the ratio of their capacitances.

##C_{1}:C_{2} = 2:3##, so charge on plate ##D = -36 \mu C##, and charge on plate ##A = -24 \mu C##.

Am I correct so far?

I think that's right.

Ok, but I am confused about one aspect of my solution.

The voltage drop across capacitor ##CD## going from ##C## to ##D## must be equal to the voltage drop across capacitor ##BA## going from ##B## to ##A##. Also, plate ##C## is at a higher potential than plate ##D##. Doesn't that mean that plate ##B## must be at a higher potential than plate ##A##?

stevendaryl
Staff Emeritus
Ok, but I am confused about one aspect of my solution.

The voltage drop across capacitor ##CD## going from ##C## to ##D## must be equal to the voltage drop across capacitor ##BA## going from ##B## to ##A##. Also, plate ##C## is at a higher potential than plate ##D##. Doesn't that mean that plate ##B## must be at a higher potential than plate ##A##?

Yes. Any two points connected by a conductor (no resistance) are at the same potential, so B&C are the same potential, and A&D are at the same potential.

In that case, isn't plate ##A## positively charged before it was connected to the capacitor ##CD##, but afterwards, plate ##A## became negatively charged due to distribution of charge between the plates ##A## and ##D##?

And similarly, isn't plate ##B## negatively charged before it was connected to the capacitor ##CD##, but afterwards, plate ##B## became positively charged due to distribution of charge between the plates ##B## and ##C##?