Electrostatics | JEE Main Solution

JEE MAIN 2025

24-01-2025 SHIFT-1

Question

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 potential energy stored in the capacitor is

Select the correct option:

$\frac{1}{4}{\varepsilon _0}{{\rm{E}}^2}{\rm{Ad}}$

$\frac{3}{4}{\varepsilon _0}{{\rm{E}}^2}{\rm{Ad}}$

${\varepsilon _0}{{\rm{E}}^2}{\rm{Ad}}$

$\frac{1}{2}{\varepsilon _0}{{\rm{E}}^2}{\rm{Ad}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution