A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is :

Solution

Physics sir · 4 steps

Energy density depends only on the field between the plates, and here E = V/d gives it directly.

1. Field between plates. The field in a parallel plate capacitor is uniform, so E = V/d = 20/(1 × 10⁻⁶) = 2 × 10⁷ V m⁻¹.

   u = 1/2ε₀ E²

2. Put in numbers. E² = 4 × 10¹⁴, so u = 1/2 × 8.85 × 10⁻¹² × 4 × 10¹⁴ = 1770 J m⁻³.
3. Cross-check. Total energy is 1/2CV² = 2 × 10⁻⁴ J. The plate area is A = Cd/ε₀ ≈ 0.113 m², so the volume is Ad ≈ 1.13 × 10⁻⁷ m³. Dividing gives the same 1770 J m⁻³.
4. So. The energy density is u ≈ 1.8 × 10³ J m⁻³, which is option (a).

The idea to keep. Energy density is 1/2ε₀ E², energy per unit volume. For parallel plates find E = V/d first. Do not stop at 1/2CV², because that is the total energy.

Answer from the NTA final answer key. Solution by PrepWiser.