Electrostatics Question 116
Question: A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V0. 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 [Kerala PET 2002]
Options:
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} $
Show Answer
Answer:
Correct Answer: D
Solution:
Work done $ W=U _{f}-U _{i} $
$ U _{i}=\frac{1}{2}CV _{0}^{2}\text{ and }{{\text{U}} _{\text{f}}}=\frac{1}{2}\frac{(C)}{3}.{{(3V _{0})}^{2}} $
$ =3\times \frac{1}{2}CV _{0}^{2} $ So $ W=\frac{{\varepsilon _{0}}AV _{0}^{2}}{d} $
BETA
