Problem 90 - A Complete Resource Book in Physics for JEE Main

Question

In a parallel plate capacitor of capacitance \(\mathrm{C}\), a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plate. The capacitance now becomes (A) \(4 \mathrm{C}\) (B) \(2 \mathrm{C}\) (C) \(\mathrm{C} / 2\) (D) \(\mathrm{C} / 4\)

Step-by-Step Solution

Answer

The capacitance of the parallel plate capacitor after inserting the metal sheet becomes \(2C\), as the metal sheet divides the capacitor into two smaller capacitors connected in series, each with capacitance \(2C\). So, the correct answer is (B) \(2C\).

Step 1: Understand the setup

A parallel plate capacitor has two conductor plates with a dielectric medium between them. The capacitance C of the capacitor depends on the area A of the plates, the separation between the plates d, and the dielectric constant K of the medium between the plates, using the formula: \[C = K \cdot\epsilon_0\cdot\frac{A}{d}\] where \(\epsilon_0\) is the vacuum permittivity constant. Now, a metal sheet with thickness half of the separation between the plates, has been inserted between the plates, parallel to them.

Step 2: Determine the effect of the metal sheet on capacitance

Since the metal sheet is a conductor and it covers the entire area between the plates, it can be considered to divide the capacitor into two smaller capacitors connected in series. Each of these capacitors has the same area A as the original one, but the separation between their plates is now half of the original separation (d/2).

Step 3: Find the capacitance of each smaller capacitor

For each smaller capacitor, the new separation between their plates is d/2. We can use the capacitance formula to find the capacitance of each of the new capacitors: \[C' = K \cdot\epsilon_0\cdot\frac{A}{d/2}= 2K\epsilon_0\cdot\frac{A}{d}\] As the dielectric constant K and vacuum permittivity \(\epsilon_0\) remain the same, we can express the capacitance of each smaller capacitor in terms of the original capacitance: \[C' = 2C\]

Step 4: Determine the new capacitance of the whole setup

Since the two smaller capacitors are connected in series, their total capacitance C'' can be found using the formula for series capacitors: \[\frac{1}{C''} = \frac{1}{C'} + \frac{1}{C'}\] Let's plug in the value of C': \[\frac{1}{C''} = \frac{1}{2C} + \frac{1}{2C}\] Simplifying this equation: \[\frac{1}{C''} = \frac{2}{2C}\] Now, take the reciprocal of both sides to find the new capacitance C'': \[C'' = 2C\]

Step 5: Match with the given options

Finally, we can compare our result with the options given: (A) 4C (B) 2C (C) C/2 (D) C/4 The correct answer is (B) 2C.