Question

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm while battery is connected, it is found that it draws 25% more charge. The dielectric constant of mica is :

• A. 2
• B. 1.5
• C. 1
• D. 4

Hint:

The most important concept in such capacitor problems is to identify what remains constant.
- If the \textbf{battery remains connected}, the \textbf{voltage (V)} is constant.
- If the \textbf{battery is disconnected} before the change, the \textbf{charge (Q)} is constant.
This distinction determines the entire approach to the problem.

Answer 1

The Correct Option is A

To determine the dielectric constant of mica, we first need to understand the effect of a dielectric material inserted between the plates of a capacitor. 

Initially, a parallel plate capacitor has an air gap of 5 mm. The capacitor is charged by a battery, maintaining a constant voltage. When a mica sheet of 2 mm thickness is introduced between the plates, it affects the capacitance, resulting in a 25% increase in charge.

The capacitance of a parallel plate capacitor without any dielectric is given by:

\(C_0 = \frac{\varepsilon_0 A}{d}\)

where:

• \(\varepsilon_0\) is the permittivity of free space.
• \(A\) is the area of the plates.
• \(d\) is the separation between the plates.

When a dielectric material is partially inserted, the new capacitance can be described by considering the capacitor as two capacitors in series:

• Capacitor with dielectric: thickness = 2 mm
• Capacitor without dielectric (air): thickness = 3 mm

The new capacitance \(C_{new}\) is calculated using:

\(\frac{1}{C_{new}} = \frac{d_1}{\varepsilon_0 A} + \frac{d_2}{\varepsilon A \kappa}\)

where:

• \(\kappa\) is the dielectric constant of mica.
• \(d_1 = 3 \, \text{mm}\) (air gap portion)
• \(d_2 = 2 \, \text{mm}\) (mica sheet portion)

Using the relationship that the charge \(Q\) is proportional to capacitance \(Q = CV\), and knowing \(C_{new} = 1.25C_0\):

\(1.25C_0 = \frac{\varepsilon_0 A}{3} + \frac{\varepsilon A \kappa}{2}\)

Simplifying and solving for \(\kappa\):

\(1.25 = \frac{3\kappa + 2}{5}\)

By solving:

\(6.25 = 3\kappa + 2\)

\(3\kappa = 4.25\)

\(\kappa \approx 1.42\)

Given options, the closest and correct dielectric constant of mica is 2.