Capacitor and Dielectric [closed]

Here's a homework problem:

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 $x_1$.

Here's my attempt at the 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 $-30 \mu C$.

closed as unclear what you're asking by Bill N, user36790, ACuriousMind♦, Gert, CuriousOneApr 14 '16 at 4:25

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• What is your question? – Bill N Apr 13 '16 at 15:50

(a)You need to have the potential difference across both capacitors to be the same so as their capacitances are different so must be the charges on their plates.

So think about $V= \frac QC$ to decide how the net charges are distributed.

(b) $C_2$ can be thought of a combination of two capacitors in series and that combination then being in parallel with $C_1$

(c) To save on the subscripts I have called $x_1$ as given in the question, $x$.
In this problem you must not forget about the battery which provides a constant potential difference $V$ across the capacitor plates.
As the dielectric is pulled out the capacitance of the capacitor drops so to maintain a constant potential difference work must be done on the battery to move the charge.

The reduction in energy stored $-dU$ in the capacitor $C$ when the dielectric is moved out a distance $dx$ in a direction parallel to the sides of the plates is given by

$$-dU = Fdx - VdQ$$

where $F$ is the force on the dielectric and $dQ$ is the charge that flows through the battery and $VdQ$ is the work done on the battery.

So $F = -\frac{dU}{dx} + V\frac{dQ}{dx}$ with $U= \frac 1 2 CV^2$ and $Q=CV \Rightarrow F = \frac 1 2 V^2 \frac {dC}{dx}$.

Knowing an expression for $C$ in terms of $x$ the force $F$ can be found.

• The electric force doesn't do work on a dielectric, does it? I would think that the work-energy gained by reducing the capacitance would go into heating the load on the battery. – sig_seg_v Apr 13 '16 at 13:51
• The dielectric is attracted into the capacitor so work must be done to pull it out. I have just found this answer which mine in part duplicates. physics.stackexchange.com/questions/248432/… – Farcher Apr 13 '16 at 13:56

(a) Of course after redistribution of charges among plates $A$ and $D$, an excess negative charge of $60\mu C$ will remain on the wire connecting plates $A$ and $D$.

But remaining charge of D is not $30\mu C$.

You have $V_C-V_D=V_B-V_A$ and $12\mu F (V_C-V_D )+ 18\mu F (V_B-V_A)=60\mu C$.

So charge of plate $D$ is $-\frac{18\mu F}{ 12\mu F+ 18\mu F } 60\mu C=-36\mu C$.

(b) $\frac {1}{C_1}=\frac {d}{\epsilon x_1^2}$ and $\frac {1}{C_2}=\frac {d-t}{\epsilon x_1^2}+\frac {t}{\epsilon _r x_1^2}.$

So, $\frac {C_1}{C_2}=\frac {d-t}{d}+\frac {\epsilon}{\epsilon _r}\frac {t}{d}=1-\frac {t(\epsilon_r- \epsilon )}{d\epsilon_r}$.

Therefore, $\frac {t}{d}=1-\frac{C_1}{C_2}\frac {\epsilon_r}{\epsilon_r-\epsilon}$.

(c) I can't understand your question exactly.