Question

Two metal plate form a parallel plate capacitor. The distance between the plates is \(\mathrm{d}\). A metal sheet of thickness \(\mathrm{d} / 2\) and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (A) \(4: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) \(5: 1\)

Short answer

The ratio of the capacitance in the two cases is (C) \(2:1\).

Step-by-step solution

Understand the capacitance formula for parallel plate capacitor

Capacitance (C) between two plates is given by the formula: \(C = \frac{\epsilon A}{d}\) Where ε is the permittivity of the dielectric material between the plates, A is the area of the plates, and d is the distance between them.

Capacitance in the first case (C1)

The first case, the distance between the plates is d. Using the capacitance formula for a parallel plate capacitor: \(C_1 = \frac{\epsilon A}{d}\)

Capacitance in the second case (C2)

In the second case, a metal sheet is introduced between the plates and it has a thickness of d/2. The metal sheet divides the capacitor into two capacitors connected in series, each with a thickness of d/2. The new capacitance (C2) is equal to the capacitance of one of these new capacitors multiplied by 2 in series: \(C'_2 = \frac{\epsilon A}{d/2}\) As the two capacitors are in series, their total capacitance (C2) can be found using the formula for capacitors in series: \(\frac{1}{C_2} = \frac{1}{C'_2} + \frac{1}{C'_2}\) Which can be simplified to: \(\frac{1}{C_2} = 2 \cdot \frac{1}{C'_2}\)

Determine the ratio of capacitance

We now want the ratio \(\frac{C_1}{C_2}\). Let's plug in the values we found for C1 and C2: \(\frac{C_1}{C_2} = \frac{\frac{\epsilon A}{d}}{\frac{1}{2 \cdot \frac{1}{\frac{\epsilon A}{d/2}}}}\) Simplifying: \(\frac{C_1}{C_2} = \frac{2 \epsilon A d}{d \epsilon A}\) Which yields: \(\frac{C_1}{C_2} = \frac{2}{1}\) Hence, the capacitance ratio in the two cases is 2:1. The correct answer is (C) \(2:1\).