Question

A parallel plate capacitor with plate area A and plate separation d is filled with a dielectric material of dielectric constant K = 4. The thickness of the dielectric material is x, where x < d. Let C1 and C2 be the capacitance of the system for \(x =\) \(\frac{1}{3}d\) and \(x=\frac{2}{3}d\), respectively. If C₁ = 2μF, the value of C₂ is _______ μF.

Answer 1

Correct Answer: 3

\(\frac{1}{C_{eq}}=\frac{1}{C_{air}}+\frac{1}{C_{di}}\)
\(\frac{1}{C_{eq}}=\frac{2d}{3(\epsilon _{0}A)}+\frac{d}{(3)4\epsilon _{0}A}=\frac{3d}{4A\epsilon _{0}}\)
\(C_{eq}=\frac{4A\epsilon _{0}}{3d}=2\mu F\)
\(\frac{A\epsilon _{0}}{d}=1.5\mu F\)
\(\frac{1}{C'_{eq}}=\frac{d}{3(\epsilon _{0}A)}+\frac{2d}{(3)4\epsilon _{0}A}\)= \(\frac{(4+2)d}{12\epsilon _{0}{A}}\)\(=\frac{6d}{12\epsilon _{0}{A}}\)\(\Rightarrow C'_{eq}=2[\frac{\epsilon _{0}A}{d}]=3\mu F\)

So, the correct answer is 3 μF

Answer 2

Capacitance Calculation with Dielectric Slab 

Step 1: Capacitance Formula for a Capacitor with a Dielectric Slab

When a dielectric slab of thickness \( x \) and dielectric constant \( K \) is inserted between the plates of a capacitor with plate separation \( d \), the system can be considered as two capacitors in series: one with the dielectric (\( C_a \)) and one with air (\( C_b \)). The formula for the equivalent capacitance is:

\( \frac{1}{C} = \frac{1}{C_a} + \frac{1}{C_b} = \frac{d - x}{\varepsilon_0 A} + \frac{x}{K\varepsilon_0 A} \)

\( C = \frac{\varepsilon_0 A}{d - x} \left( 1 + \frac{x}{K} \right) \)

Step 2: Capacitance \( C_1 \)

Given \( x = \frac{1}{3}d \) and \( K = 4 \), the capacitance \( C_1 \) is calculated as follows:

\( C_1 = \frac{\varepsilon_0 A}{d - \frac{d}{3}} \left( 1 + \frac{\frac{d}{3}}{4} \right) = \frac{\varepsilon_0 A}{\frac{2d}{3}} \left( 1 + \frac{d}{12} \right) \)

Simplifying further:

\( C_1 = \frac{\varepsilon_0 A}{\frac{2d}{3}} \left( \frac{12 + d}{12} \right) = \frac{12\varepsilon_0 A}{9d} = \frac{4\varepsilon_0 A}{3d} \)

Step 3: Capacitance \( C_2 \)

Given \( x = \frac{2}{3}d \) and \( K = 4 \), the capacitance \( C_2 \) is calculated as follows:

\( C_2 = \frac{\varepsilon_0 A}{d - \frac{2d}{3}} \left( 1 + \frac{\frac{2d}{3}}{4} \right) = \frac{\varepsilon_0 A}{\frac{d}{3}} \left( 1 + \frac{d}{6} \right) \)

Simplifying further:

\( C_2 = \frac{\varepsilon_0 A}{\frac{d}{3}} \left( \frac{6 + d}{6} \right) = \frac{6\varepsilon_0 A}{3d} = \frac{2\varepsilon_0 A}{d} \)

Step 4: Relationship between \( C_1 \) and \( C_2 \)

We have \( C_1 = \frac{4\varepsilon_0 A}{3d} \) and \( C_2 = \frac{2\varepsilon_0 A}{d} \). We can express \( C_2 \) in terms of \( C_1 \):

\( C_2 = \frac{2\varepsilon_0 A}{d} = \frac{3}{2} \cdot \frac{4\varepsilon_0 A}{3d} = \frac{3}{2} C_1 \)

Step 5: Value of \( C_2 \)

Given \( C_1 = 2 \, \mu F \), we can calculate \( C_2 \):

\( C_2 = \frac{3}{2} \times 2 \, \mu F = 3 \, \mu F \)

Conclusion:

The value of \( C_2 \) is \( \mathbf{3 \, \mu F} \).