The material filled between the plates of a parallel plate capacitor has resistivity 200 Ωm. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is : (given the value of relative permittivity of material is 50)

Solution

Physics sir · 4 steps

The same slab sets both R and C, so RC = ρ Kε₀ and the plate geometry drops out.

1. Two formulas, same plates. The filling conducts a little, so it acts as a resistor R = (ρ d)/A between the plates. The same slab gives the capacitance C = (Kε₀ A)/d.

   RC = (ρ d)/A × (Kε₀ A)/d = ρ K ε₀

2. Find R. RC = 200 × 50 × 8.85 × 10⁻¹² = 8.85 × 10⁻⁸ s. Dividing by C = 2 × 10⁻¹² F gives R = 4.425 × 10⁴ Ω.
3. Leakage current. I = V/R = 40/(4.425 × 10⁴) ≈ 9.04 × 10⁻⁴ A.
4. So. The leakage current is about 0.9 mA, which is option (c).

The idea to keep. A leaky capacitor is a resistor and a capacitor built from the same slab. Multiply them and you get RC = ρ Kε₀, so you never need the area or the plate gap.

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