What is the mechanical work done in pulling the slab out of the capacitor if the battery is connected
Answer & explanation
Correct answer: option 1
Work done = change in potential energy = U2 – U1
$ U_1 = \frac{1}{2} kCE^2$
$U_2 = \frac{1}{2}CE^2 $
$\Rightarrow W = U_2 - U_1 = \frac{1}{2} CE^2(k-1)$