A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness (1/3)ʳᵈ of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

Solution

Physics sir · 4 steps

Treat the slab and the leftover air gap as two capacitors in series, then compare with the vacuum value.

1. Two layers in series. With vacuum, C = (ε₀ A)/d. The sheet of thickness t = d/3 and the remaining air gap 2d/3 sit one after the other between the plates, so they act in series. The slab behaves like an air gap of thickness t/K.

   C' = (ε₀ A)/((d - t) + t/K)

2. Put in numbers. The denominator is 2d/3 + d/3K = d(2K+1)/3K. So C' = 3K/(2K+1) × (ε₀ A)/d = 3KC/(2K+1).
3. Quick check. Put K = 1, which means the sheet is just vacuum. Then 3KC/(2K+1) = 3C/3 = C, the original value, as it must be.
4. So. The new capacitance is 3KC/(2K+1), which is option (d).

The idea to keep. A dielectric slab of thickness t acts like an air gap of only t/K. Replace d by d - t + t/K, then test your answer with K = 1.

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