Capacitor and Dielectric: Charge, Arrangement, and Extraction Calculations

[spaghetti3451]

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 $x_1$.

Homework Equations

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]

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 $x_1$.

Homework Equations

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 $D$?

 

[spaghetti3451]

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]

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.

 

[spaghetti3451]

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]

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.

 

[spaghetti3451]

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 $C$?