Question

A parallel plate capacitor has plate area $A$ and separation $d$. It is charged to a potential difference $V_0$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is:

• A. $\frac{3{ϵ}_{0}A{V}_0^2}{d}$
• B. $\frac{{ϵ}_{0}A{V}_0^2}{2d}$
• C. $\frac{{ϵ}_{0}A{V}_0^2}{3d}$
• D. $\frac{{ϵ}_{0}A{V}_0^2}{d}$

Answer 1

Answer: (D)

$\frac{{ϵ}_{0}A{V}_0^2}{d}$

Here, $W=U_f-U_i$

where, $U_i=\frac{1}{2}C{V}_0^2$

$U_f=\frac{1.C.(3V_0{)}^2}{2.3}$

=$3\ast \frac{1}{2}C{V}_0$

So, $W=\frac{{ϵ}_{0}A{V}_0^2}{d}$