Net energy loss in infinite series of capacitors

1 vote
35 views 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?

https://imgur.com/PqsK79D edited May 16

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 $$W-E_{\infty}=\frac{5Q^2}{6C}=\frac56CV^2=\frac{20}{6}\pi\epsilon_0 RV^2$$