Question

The distance between two plates of a capacitor is \(d\) and its capacitance is \(C_1\), when air is the medium between the plates. If a metal sheet of thickness \(\frac {2d}{3}\) and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes \(C_2\). The ratio \(\frac {C_2}{C_1}\) is

• A. 4:1
• B. 2:1
• C. 3:1
• D. 1:1

Answer 1

The Correct Option is C

The capacitance \( C \) of a parallel plate capacitor 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. For the initial configuration, the capacitance is \( C_1 \) with air as the dielectric, so: \[ C_1 = \frac{\epsilon_0 A}{d} \] When a metal sheet of thickness \( \frac{2d}{3} \) is introduced between the plates, the effective distance between the plates becomes \( d - \frac{2d}{3} = \frac{d}{3} \), and the capacitance becomes \( C_2 \). The formula for \( C_2 \) becomes: \[ C_2 = \frac{\epsilon_0 A}{d - t + \frac{t}{K}} \] where \( t = \frac{2d}{3} \) and \( K = \infty \) for metals. Substituting these values: \[ C_2 = \frac{\epsilon_0 A}{\frac{d}{3}} = 3 \times \frac{\epsilon_0 A}{d} = 3 C_1 \] Thus, the ratio \( \frac{C_2}{C_1} = 3 : 1 \).