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Finding the current in the capacitor circuit

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I am confused about the initial step of this problem. I know that charge on $C_1$ will decrease and that on $C_2$ will increase but it by the same amount? Really unable to figure out that particular part. The answerer may provide a hint so that I can begin and if I have any further questions, I will ask them through comments.

asked Jul 18 in Physics Problems by Reststack (404 points)

1 Answer

2 votes

Everything you need to solve this problem has arisen in the other capacitor problems you have worked on recently.

The starting point is to look for formulae which connect the separation of the plates $x$, the capacitance $C$, the charge stored $Q$, and the potential difference $V$. Note that (i) the total charge $Q_1+Q_2$ on the 2 capacitors is fixed, and (ii) the PD across the capacitors is always the same : $V_1=V_2$.

The charge on $C_1$ will increase, not decrease : the plates are getting closer so the capacitance increases and the charge increases with it. Vice versa for $C_2$.

The speed of the plates is $\frac{dx}{dt}$. The current leaving $C_2$ and arriving on $C_1$ is $\frac{dQ_1}{dt}=-\frac{dQ_2}{dt}$ (because $Q_1$ is increasing).


The total charge on the two capacitors is fixed : $$Q_1+Q_2=Q_0=400\mu C$$ The capacitance of each capacitor is $C=\frac{\epsilon A}{x}$. The voltage across each capacitor changes as the separation between the plates changes, but the two voltages are always equal so $$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$$ $$Q_1 x_1=Q_2 x_2$$ $$\dot Q_1x_1+Q_1\dot x_1=\dot Q_2x_2+Q_2\dot x_2$$ $$\dot Q_1(x_1+x_2)=-\dot x_1(Q_2+Q_1)=-\dot x_1 Q_0$$ in which I have used the fact that $\dot Q_2=-\dot Q_1$ and $Q_2=Q_0-Q_1$ and $\dot x_2=-\dot x_1$.

It is not clear at what moment the current in the circuit is to be evaluated. I shall assume it is $t=0$. Then $x_1+x_2=0.2m$ and $\dot x_1=-0.001 m/s$ so the current in the circuit is $$\dot Q_1(0)=\frac{0.001m/s}{0.2m}\times 400\mu C=2.0A$$

In fact $x_1+x_2=\text{constant}$ so this is the constant current which flows until the plates of $C_2$ touch at $t=100s$ when $C_1$ discharges because there is a short-circuit across it.

answered Jul 18 by sammy gerbil (26,482 points)
edited Jul 21 by sammy gerbil
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