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

To determine the leakage current flowing out of the capacitor, we need to first understand a few important concepts and relationships.

1. Ohm's Law: It relates the current \(I\), voltage \(V\), and resistance \(R\) in a circuit as follows: \(I = \frac{V}{R}\).
2. Resistance and Resistivity: The resistance \(R\) is related to resistivity \(\rho\), length \(L\), and cross-sectional area \(A\) as: \(R = \frac{\rho \cdot L}{A}\).
3. Capacitor Geometry: For a parallel plate capacitor, if \(C\) is the capacitance, then for a dielectric material inserted: \(C = \varepsilon_r \varepsilon_0 \frac{A}{d}\), where \(\varepsilon_r\) is the relative permittivity, \(\varepsilon_0\) is the vacuum permittivity, \(A\) is the plate area, and \(d\) is the separation distance between the plates. 

Given:

• Capacitance \(C = 2 \, \text{pF} = 2 \times 10^{-12} \, \text{F}\)
• Relative permittivity \(\varepsilon_r = 50\)
• Resistivity \(\rho = 200 \, \Omega\text{m}\)
• Voltage \(V = 40 \, \text{V}\)

 

From the capacitance formula, solve for \(R\):

1. Knowing \(C = \varepsilon_r \varepsilon_0 \frac{A}{d}\), and that \(d = \frac{\varepsilon_r \varepsilon_0 A}{C}\), recognize that from \(R = \frac{\rho \cdot d}{A}\), substitute for \(d\) using: \(d = \frac{\varepsilon_r \varepsilon_0 A}{C}\), and consequently: \(R = \frac{\rho \cdot \varepsilon_r \varepsilon_0}{C}\).
2. Calculate:
   • \(\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}\)
   • \(\varepsilon_r = 50\)
3. Apply Ohm’s Law: \(I = \frac{V}{R} = \frac{40}{4427} \approx 0.009 \, \text{A} = 0.9 \, \text{mA}\).

 

Therefore, the leakage current is approximately 0.9 mA, which matches the correct option provided.