Q 12-02-055JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
Consider a parallel plate capacitor of area $A$ (of each plate) and separation $d$ between the plates. If $E$ is the electric field and $\varepsilon_0$ is the permittivity of free space between the plates, then the potential energy stored in the capacitor is
Answer: (B) $\dfrac{1}{2}\varepsilon_0E^2Ad$
Energy density in the field is $u = \tfrac{1}{2}\varepsilon_0E^2$, and the field fills volume $Ad$:
$$U = \frac{1}{2}\varepsilon_0E^2Ad$$
(Check: $\tfrac{1}{2}CV^2 = \tfrac{1}{2}\dfrac{\varepsilon_0A}{d}(Ed)^2$ gives the same.)
Solution by Sreeraj P, M.Sc Physics