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Net energy loss in infinite series of capacitors

1 vote

Here $Q=4\pi \epsilon_0 RV$ and $ C=4\pi \epsilon_0 R$

While solving i got the energy loss when S2 is closed as $$E_i-E_f = \frac{Q^2}{2C} - (\frac{Q^2}{8C} + \frac{Q^2}{8C})$$
since $Q$ on 1st capacitor equally divides between 1st and 2nd capacitor.

So energy loss when S2 is closed is $\frac{Q^2}{4C}$ and that when S3 is closed is $\frac{Q^2}{16C}$ and so on.

When I take the sum I get $\frac43\pi\epsilon_0 RV^2$ but that's not the answer.

What went wrong?


asked May 16, 2019 in Physics Problems by qwerty (110 points)
edited May 16, 2019 by sammy gerbil

1 Answer

0 votes

You have forgotten that there is work done by the battery after $t=0$, so that there is energy lost before S1 is opened and S2 is closed.

The work done by the battery to transfer charge $Q$ onto the 1st capacitor is $W=QV=\frac{Q^2}{C}$.

The final charges on the capacitors are $\frac12Q, \frac14Q, \frac{1}{16}Q, \frac{1}{64}Q, ...$ The total energy stored in all the capacitors at this time is $$E_{\infty}=\frac{Q^2}{2C}[(\frac12)^2+(\frac14)^2+(\frac{1}{16})^2+(\frac{1}{64})^2 +...]=\frac{Q^2}{6C}$$ The total loss of energy after closing S1 is $$W-E_{\infty}=\frac{5Q^2}{6C}=\frac56CV^2=\frac{20}{6}\pi\epsilon_0 RV^2$$

So I think option A should be the correct answer. The marking scheme appears to be wrong.

answered May 16, 2019 by sammy gerbil (28,896 points)