Question

An air-filled capacitor with plate area A and plate separation d has capacitance \( C_0 \). A slab of dielectric constant K, area A and thickness \( \frac{d}{5} \) is inserted between the plates. The capacitance of the capacitor will become:

• A. \( \left[ \frac{4K}{5K+1} \right] C_0 \)
• B. \( \left[ \frac{K+5}{4} \right] C_0 \)
• C. \( \left[ \frac{5K}{4K+1} \right] C_0 \)
• D. \( \left[ \frac{K+4}{5K} \right] C_0 \)

Hint:

When a dielectric is inserted into a capacitor, the capacitance increases by a factor of the dielectric constant. The capacitance is determined by the ratio of the thickness of the dielectric region to the total plate separation.

Answer 1

The Correct Option is C

Step 1: The capacitance of a parallel plate capacitor without any dielectric is given by: \[ C_0 = \epsilon_0 \frac{A}{d} \] where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates. Step 2: When a dielectric material with dielectric constant \( K \) is inserted between the plates, the capacitance increases by a factor of \( K \), but only for the portion of the plate that is covered by the dielectric. In this case, the dielectric slab occupies \( \frac{d}{5} \) of the total plate separation \( d \). Step 3: The new capacitance is given by the sum of the capacitance of the dielectric-filled region and the air-filled region: \[ C = \frac{\epsilon_0 A}{d - \frac{d}{5}} + \frac{\epsilon_0 K A}{\frac{d}{5}} \] Simplifying this expression: \[ C = \frac{\epsilon_0 A}{\frac{4d}{5}} + \frac{\epsilon_0 K A}{\frac{d}{5}} = \frac{5\epsilon_0 A}{4d} + \frac{5K \epsilon_0 A}{d} \] Step 4: Now, factoring out the common term \( \frac{\epsilon_0 A}{d} \), we get: \[ C = \frac{\epsilon_0 A}{d} \left( \frac{5}{4} + 5K \right) = C_0 \left( \frac{5}{4} + 5K \right) \] Step 5: Therefore, the new capacitance is: \[ C = C_0 \left[ \frac{5K}{4K+1} \right] \] Thus, the capacitance of the capacitor becomes \( \left[ \frac{5K}{4K+1} \right] C_0 \).