JEE Main 2022 · 28 July · evening · Single correct · +4 / −1
A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 3/4 d, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :
(Given C₀ = capacitance of capacitor with air as medium between plates.)
- (a) (4 K C₀)/(3+K)
- (b) (3 K C₀)/(3+K)
- (c) (3+K)/(4 K C₀)
- (d) K/(4+K)
Answer and solution
Answer (a) (4 K C₀)/(3+K)
Solution
Physics sir · 3 steps
The slab only changes the gap the field crosses: air counts fully, dielectric thickness counts as t/K.
- Effective gap. A dielectric slab of thickness t weakens the field inside it by a factor K. So that thickness acts like an air gap of only t/K, while the remaining air gap d - t counts in full.
C = (ε₀ A)/(d - t + t/K)
- Put in numbers. With t = 3/4d, the denominator is d/4 + 3d/4K = d(K+3)/4K. So C = 4K/(K+3) × (ε₀ A)/d = 4KC₀/(3+K).
- So. The capacitance is 4KC₀/(3+K), option (a). Check: with K = 1 (no real dielectric) it gives C₀, as it should.
The idea to keep. For a partly filled capacitor, replace each dielectric thickness t by t/K in the gap. Then test your answer at K = 1; it must give back C₀.
Answer from the NTA final answer key. Solution by PrepWiser.