Question

Two identical sheets of the same metallic foil of the area $A$ are separated by a distance $d$, and charged by a battery of emf $E$. Keeping the charge constant, the separation is increased by $l$. Then, the new capacitance and potential difference will be

• A. $\frac{{ϵ}_{0}A}{d},E$
• B. $\frac{{ϵ}_{0}A}{(d+l)},E$
• C. $\frac{{ϵ}_{0}A}{(d+l)},[1+\frac{l}{d}]E$
• D. $\frac{{ϵ}_{0}A}{d},[1+\frac{l}{d}]E$

Answer 1

The correct option is C $\frac{{ϵ}_{0}A}{(d+l)},[1+\frac{l}{d}]E$

Initial capcitance $C$ will be

$C=\frac{A{ϵ}_0}{d}$

If charge is kept constant,

$Q_i=Q_f.......\left(1\right)$

By using formula, $Q=CV$

$\therefore Q_i=\frac{A{ϵ}_0}{d}E$

Let the final potential difference be $E^{\prime }$.

Final capacitance $C^{\prime }$ when separation is increased by $l$ will be

$C^{\prime }=\frac{A{ϵ}_0}{d+l}$

Again using formula, $Q=CV$

$\therefore Q_f=\frac{A{ϵ}_0}{d+l}{E}^{\prime }$

putting the values in eq.$\left(1\right)$

$\left(\frac{A{ϵ}_0}{d+l}\right)E^{\prime }=\frac{A{\epsilon }_0}{d}E$

$\Rightarrow E^{\prime }=E(1+\frac{l}{d})$

Hence, option $\left(c\right)$ is the correct choice.