MHT CET Medical202623 April 2026Evening ShiftPhysicsCapacitanceActual
A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C₁ . When the capacitor is charged, the plate area covered by the dielectric gets charge Q₁ and the rest of the area gets charge Q₂ . The electric field in the d
Options
- AE₁ E₂ =4
- BE₁ E₂ = 1 K
- CC C₁ = 2+K K
- DQ₁ Q₂ = 3 K
Correct answer
C. C C₁ = 2+K K
Step-by-step solution
The given arrangement can be considered as two capacitors connected in parallel. Let the total area of the plates be A and the distance between them be d . The capacitance of the portion with the dielectric (covering 1/3 of the area) is: C₁ = K ₀ (A/3) d = K ₀ A 3d The capacitance of the remaining portion (covering 2/3 of the area) is: C₂ = ₀ (2A/3) d = 2 ₀ A 3d The total capacitance C is the sum of C₁ and C₂ since they are in parallel: C = C₁ + C₂ = K ₀ A 3d + 2 ₀ A 3d = ₀ A 3d (K + 2) Now, finding the ratio C C₁