A parallel plate capacitor with air between the plate has a capacitance of 15pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes x/4 pF. The value of x is ____________.

Solution

Physics sir · 3 steps

Capacitance goes up by K and down by the gap, so multiply 15 pF by 3.5 and halve it.

1. Start from the formula. With air, C = (ε₀ A)/d = 15 pF. A dielectric multiplies this by K, and a larger gap divides it.

   C' = (Kε₀ A)/2d = K/2×(ε₀ A)/d = K/2 C

2. Put in numbers. C' = 3.5/2 × 15 = 26.25 pF. Written over 4, this is 105/4 pF.
3. So. x/4 = 105/4, so x = 105.

The common slip: 210. Forgetting that the gap doubled gives C' = 3.5 × 15 = 52.5 pF, which makes x = 210, not 105.

The idea to keep. For a parallel plate capacitor, C ∝ KA/d. Change each factor as a ratio and multiply the old capacitance by it; you never need ε₀ or A themselves.

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