Verified Solution • 2025–26 Session moderate

QJEEM25Ph1c6b: JEE Main 2025 Physics - Class 12 JEEM25Ph1c6b Solution: Consider a parallel plate capacitor of area A (of each plate)…

Consider a parallel plate capacitor of area A (of each plate) and separation ' dd ' between the plates. If EE is the electric field and ε0\varepsilon_0 is the permittivity of free space between the plates, then potential energy stored in the capacitor is

Multiple Choice Options

  • A 14ε0E2Ad\frac{1}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}
  • B 12ε0E2Ad\frac{1}{2} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}
  • C ε0E2Ad\varepsilon_0 \mathrm{E}^2 \mathrm{Ad}
  • D 34ε0E2Ad\frac{3}{4} \varepsilon_0 \mathrm{E}^2 \mathrm{Ad}

Step-by-Step Verified Solution

u=12ε0E2

The energy density of an electric field in free space is given by the expression above. For a parallel plate capacitor with plate area A and separation d, the volume between the plates is

V=Ad.

Multiplying the energy density by the volume gives the total energy stored:

U=u×V=12ε0E2×Ad=12ε0E2Ad.

Thus, the potential energy stored in the capacitor is

12ε0E2Ad.

This corresponds to the option:

Option D: 12ε0E2Ad.

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