Question

A capacitor of capacitance $C$ is charged by connecting it to a battery of emf $E$. The capacitor is now disconnected and reconnected to the battery with the polarity reversed. The heat developed in the connecting wires is:

• A. $2C{E}^2$
• B. $C{E}^2$
• C. $\frac{C{E}^2}{4}$
• D. $\frac{3}{2}C{E}^2$

Answer 1

The correct option is A $2C{E}^2$

When capacitor is connected to the battery of emf $E$, the charge $q$ of capacitor will be

$q=CE$


After the battery is disconnected, charge on the capacitor remains the same.


Situation after reconnecting battery with opposite polarity or reversed polarity


We can see that the left plate of capacitor which was initially positively charged has now become negatively charged with charge $q^{\prime }$.

$\therefore q^{\prime }=CV=CE$

In final situation charge on left plate is $-CE$ and initially it was $+CE$.

Therefore, charge transferred by the battery is $-2CE$ to the left plate.

So, work done $W$ by battery is given by

$W=\left(2CE\right)\left(E\right)=2C{E}^2$

$[\because W=qV$ and magnitude of charge is $2CE]$

Heat developed in wires is given by

Work done by battery $=$ Change in potential energy of the capacitor $+$ energy loss in the form of heat

$\therefore H+(U_f-U_i)=W_{Battery}$

$(\because U_f=U_i=\frac{1}{2}C{E}^2)$

$\Rightarrow H=W_{battery}=2C{E}^2$

Therefore, option $\left(a\right)$ is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: In such problems always look for}\\ \text{charge supplied by battery. Therefore,}\\ \text{we will know the total energy supplied}\\ \text{\& work done by battery, a part of which}\\ \text{will be stored in capacitor and rest is}\\ \text{dissipated in the form of heat.}\end{array}$