The gap between the plates of a parallel plate capacitor is filled with glass of resistivity `rho`. The capacitance of the capacitor without glass equals C. The leakage current of the capacitor when a voltage V is applied to it is

Options

A. `(V rho)/(C epsilon_(0))`
B. `(CV)/(rho epsilon_(0))`
C. `(V epsilon_(0))/(C rho)`
D. `(CV rho)/(epsilon_(0))`

Answer

To find the leakage current of a parallel plate capacitor filled with glass of resistivity \( \rho \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Capacitance**:
The capacitance \( C \) of a parallel plate capacitor without any dielectric is given by the formula:
\[ C = \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the distance between the plates.

2. **Charge on the Capacitor**:
When a voltage \( V \) is applied across the capacitor, the charge \( Q \) stored in the capacitor can be expressed as:
\[ Q = C \cdot V \]

3. **Electric Field in the Capacitor**:
The electric field \( E \) between the plates of the capacitor is given by:
\[ E = \frac{V}{d} \]

4. **Current Density**:
According to Ohm's law in terms of current density, the current density \( j \) can be expressed as:
\[ j = \sigma E \]
where \( \sigma \) is the conductivity of the material. Since conductivity \( \sigma \) is the reciprocal of resistivity \( \rho \), we have:
\[ j = \frac{1}{\rho} E \]

5. **Total Current**:
The total current \( I \) flowing through the capacitor can be found by integrating the current density over the area \( A \):
\[ I = j \cdot A = \frac{1}{\rho} E \cdot A \]

6. **Substituting Electric Field**:
Substitute the expression for the electric field \( E \):
\[ I = \frac{1}{\rho} \left(\frac{V}{d}\right) A \]

7. **Relating Current to Capacitance**:
Since \( C = \frac{\epsilon_0 A}{d} \), we can rearrange this to find \( A \):
\[ A = \frac{C d}{\epsilon_0} \]

8. **Substituting for Area**:
Substitute this expression for \( A \) back into the current equation:
\[ I = \frac{1}{\rho} \left(\frac{V}{d}\right) \left(\frac{C d}{\epsilon_0}\right) \]

9. **Simplifying**:
The \( d \) cancels out:
\[ I = \frac{C V}{\rho \epsilon_0} \]

10. **Final Expression for Leakage Current**:
Thus, the leakage current \( I \) when a voltage \( V \) is applied to the capacitor is given by:
\[ I = \frac{C V}{\rho \epsilon_0} \]

### Final Answer:
The leakage current of the capacitor when a voltage \( V \) is applied is:
\[ I = \frac{C V}{\rho \epsilon_0} \]