A parallel plate capacitor with plate separation 5 mm is charged up by a battery. It is found that on introducing a dielectric sheet of thickness 2 mm, while keeping the battery connections intact, the capacitor draws 25 % more charge from the battery than before. The dielectric constant of the sheet is _________.

Solution

Physics sir · 4 steps

Battery stays connected, so V is fixed: 25% more charge means 25% more capacitance.

1. Charge to capacitance. With the battery connected, V does not change. Since Q = CV, drawing 25% more charge means C' = 1.25 C.
2. Partial dielectric. A sheet of thickness t placed in a gap d behaves like a smaller air gap. Its effective air thickness is d - t + t/K.

   C'/C = d/(d - t + t/K) = 5/(3 + 2/K) = 1.25

3. Solve for K. 3 + 2/K = 5/1.25 = 4, so 2/K = 1. The millimetres cancel in the ratio, so no unit conversion is needed.
4. So. The dielectric constant of the sheet is K = 2.

The common slip: 1.25. Treating the sheet as if it fills the whole gap, so that C' = KC, gives 1.25, not 2. The sheet covers only 2 mm of the 5 mm gap.

The idea to keep. A dielectric slab of thickness t acts like an air gap of t/K, so the effective separation is d - t + t/K. With the battery connected, the charge ratio equals the capacitance ratio.

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