Question

Consider the situation shown in figure, the switch $S$ open for a long time and then closed. Then work done by the battery will be:

• A. $\frac{C{E}^2}{2}$
• B. $\frac{C{E}^2}{4}$
• C. $2C{E}^2$
• D. $\frac{C{E}^2}{8}$

Answer 1

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

When the switch is open, the capacitors are in series. So equivalent capacitance will be

$C_{eq}=\frac{C\times C}{C+C}=\frac{C}{2}$

Charge flown by battery is

$q=C_{eq}V$

$\Rightarrow q=\frac{C}{2}E$

Thus, the same charge will be appearing on both capacitors.


After the switch $S$ is closed, the potential difference across second capacitors becomes zero instantly. Therefore, it will have zero charge or no charge.


Potential of point $\text{A}$ and $\text{B}$ is the same, since they lie on the same conducting wire.

$V_A=V_B$

$\Rightarrow \Delta V=0$

Charge on first capacitor is,

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

Therefore, charge supplied (q'') by battery will be,

$q^{\prime \prime }=q^{\prime }-q$

$\Rightarrow q^{\prime \prime }=CE-\frac{CE}{2}$

$\Rightarrow q^{\prime \prime }=\frac{CE}{2}$

Work done by battery is,

$W=\left(q^{\prime \prime }\right)\times E$

$\therefore W=\frac{CE}{2}\times E=\frac{C{E}^2}{2}$

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