Question

The material filled between the plates of a parallel plate capacitor has resistivity 200 \(\Omega\)m. 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)

• A. 9.0 mA
• B. 0.9 mA
• C. 0.9 \(\mu\)A
• D. 9.0 \(\mu\)A

Hint:

The relation \(RC = \rho \epsilon\) is extremely useful for lossy capacitor problems as it eliminates the need to know the physical dimensions (Area and distance) of the plates.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A dielectric with non-infinite resistivity allows a small current to flow through it, known as leakage current. This current follows Ohm's law \(I = V/R\), where \(R\) is the resistance of the dielectric block.
Step 2: Key Formula or Approach:
The product of Resistance (\(R\)) and Capacitance (\(C\)) of a material filling the same volume is constant:
\[ RC = \rho \epsilon = \rho \epsilon_0 \epsilon_r \]
Leakage current \(I = \frac{V}{R} = \frac{V C}{\rho \epsilon_0 \epsilon_r}\).
Step 3: Detailed Explanation:
Given:
\(V = 40 \text{ V}\)
\(C = 2 \times 10^{-12} \text{ F}\)
\(\rho = 200 \text{ } \Omega\text{m}\)
\(\epsilon_r = 50\)
\(\epsilon_0 \approx 8.85 \times 10^{-12} \text{ F/m}\)
Calculate the current:
\[ I = \frac{40 \times 2 \times 10^{-12}}{200 \times 8.85 \times 10^{-12} \times 50} \]
\[ I = \frac{80}{200 \times 50 \times 8.85} = \frac{80}{10000 \times 8.85} \]
\[ I = \frac{80}{88500} \approx 0.000903 \text{ A} = 0.9 \times 10^{-3} \text{ A} = 0.9 \text{ mA} \]
Step 4: Final Answer:
The leakage current is 0.9 mA.