Q. 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) $\frac{3\varepsilon_{0}AV_{0}^{2}}{d}$
B) $\frac{\varepsilon_{0}AV_{0}^{2}}{2d}$
C) $\frac{\varepsilon_{0}AV_{0}^{2}}{3d}$
D) $\frac{\varepsilon_{0}AV_{0}^{2}}{d}$
Explanation
(d) Work done $W = U_{f} - U_{i}$
$U_{i} = \frac{1}{2}CV_{0}^{2}$ and $U_{f} = \frac{1}{3}\frac{(C)}{3}.\left( 3V_{0} \right)^{2} = 3 \times \frac{1}{2}CV_{0}^{2}$
So $W = \frac{\varepsilon_{0}AV_{0}^{2}}{d}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.