Question

A capacitor is connected to a 10V battery. The charge on plates is 40µC when medium between plates is air. The charge on the plates become 100µC when the space between the plates is filled with oil. The dielectric constant of oil is :

• A. 2.5
• B. 4
• C. 6.25
• D. 10

Answer 1

Answer: (A)

2.5

Answer 2

The problem involves calculating the dielectric constant of oil using the charge stored on a capacitor with air and then with oil as the dielectric medium, while connected to the same battery.

1. Capacitance with air as dielectric: When the medium between the plates is air, the charge on the plates is $Q_{air} = 40 \mu C$ and the voltage across the capacitor is $V = 10 V$. The capacitance with air, $C_{air}$, is given by:

   $$C_{air} = \frac{Q_{air}}{V}$$ $$C_{air} = \frac{40 \times 10^{-6} C}{10 V} = 4 \times 10^{-6} F = 4 \mu F$$

2. Capacitance with oil as dielectric: When the space between the plates is filled with oil, the charge on the plates becomes $Q_{oil} = 100 \mu C$, and the voltage remains $V = 10 V$ (since it's connected to the same battery). The capacitance with oil, $C_{oil}$, is given by:

   $$C_{oil} = \frac{Q_{oil}}{V}$$ $$C_{oil} = \frac{100 \times 10^{-6} C}{10 V} = 10 \times 10^{-6} F = 10 \mu F$$

3. Dielectric constant of oil: The capacitance of a capacitor with a dielectric medium of dielectric constant $K$ is related to its capacitance with air ($C_{air}$) by the formula:

   $$C_{oil} = K \cdot C_{air}$$

   Therefore, the dielectric constant $K$ can be calculated as:

   $$K = \frac{C_{oil}}{C_{air}}$$ $$K = \frac{10 \mu F}{4 \mu F} = 2.5$$

The dielectric constant of oil is 2.5.