What is the mechanical work done in pulling the slab out of the capacitor after disconnecting it from the battery
Answer & explanation
Correct answer: option 1
Work done = change in potential energy = U2 – U1
v1 = (1/2) E2C
v2 = $\frac{1}{2} \frac{(E C)^2}{C'}=\frac{1}{2} \frac{E^2 C^2}{C} \varepsilon_r=(1 / 2) E^2 C \varepsilon_r$
∴ Work done = $(1 / 2) E^2 C\left(\varepsilon_r-1\right)$
∴ (A)