Question

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant \(K_1\) and \(K_2\) with thickness \(d/2\) and \(d/2\) respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If \(K_1 = 1.25 K_2\), the value of \(K_2\) is :

• A. \( 2.33 \)
• B. \( 1.60 \)
• C. \( 1.33 \)
• D. \( 2.66 \)

Hint:

Treat the capacitor with two dielectrics in series. The equivalent capacitance \(C_{eq}\) is given by \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \), where \(C_i = \frac{K_i \epsilon_0 A}{d/2} \). Use the condition \(C_{eq} = 2 C_0\) and the relation \(K_1 = 1.25 K_2\) to solve for \(K_2\).

Answer 1

The Correct Option is B

The problem involves calculating the value of the dielectric constant \(K_2\) for a parallel plate capacitor with two dielectric slabs inserted, causing the capacitance to double.

The original capacitance without dielectric is given by:

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

where \( \varepsilon_0 \) is the permittivity of free space and \(A\) is the area of the plates.

After inserting the two dielectric slabs with constants \(K_1\) and \(K_2\), the effective capacitance \(C\) becomes:

\(\frac{1}{C} = \frac{d_1}{K_1 \varepsilon_0 A} + \frac{d_2}{K_2 \varepsilon_0 A}\)

Given \(d_1 = d_2 = \frac{d}{2}\), substituting these values, we get:

\(\frac{1}{C} = \frac{d}{2K_1 \varepsilon_0 A} + \frac{d}{2K_2 \varepsilon_0 A}\)

Simplifying:

\(\frac{1}{C} = \frac{d(K_2 + K_1)}{2K_1 K_2 \varepsilon_0 A}\)

We know \(C = 2C_0\) thus:

\( \frac{1}{2C_0} = \frac{d(K_2 + K_1)}{2K_1 K_2 \varepsilon_0 A}\)

\(2K_1 K_2 = K_2 + K_1\)

Using \(K_1 = 1.25K_2\), substitute into the equation:

\(2(1.25K_2)K_2 = K_2 + 1.25K_2\)

Simplifying yields:

\(2.5K_2^2 = 2.25K_2\)

Cancel out \(K_2\) from both sides (assuming \(K_2 \neq 0\)):

\(2.5K_2 = 2.25\)

Solving for \(K_2\) gives:

\(K_2 = \frac{2.25}{2.5} = 1.60\)

Thus, the required value of \(K_2\) is 1.60.