Question

A capacitor having a capacitance of $100\mu F$ is charged to a potential difference of $24V$. The charging battery is disconnected and the capacitor is connected to another battery of emf $12V$ with the positive plate of the capacitor joined with the positive terminal of the battery. (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the $12V$ battery, (c) Is work done by the battery or is it done on the battery? Find its magnitude. (d) Find the decrease in electrostatic field energy. (e) Find the heat developed during the flow of charge after reconnection.

Answer 1

$a)$ Before reconnection: $C=100\mu F,V=24V$

The charge on capacitor is given by:

$q=CV=100\times 24=2400\mu C$

After reconnection, $C=100\mu F,V=12V$

$q=CV=1200\mu C$

$b)$ Charge through 12V battery,

$C=100\mu F,V=12V$

So, the charge will be:

$q=CV=1200\mu C$

$c)$ Work done is given as:

$W=Vq$

$\therefore W=12\times 1200=14400\mu J$,

which is positive. So, work is done by the battery.

$d)$.Decrease in electrostatic field energy

$=\frac{1}{2}C(V_1^2-V_2^2)$

$=\frac{1}{2}\times 100(576-144)=21600J$

$e)$ Heat developed=Energy appeared = $\frac{1}{2}C{V}^2$

$\frac{1}{2}\times 100\times {12}^2=7200J$