Question

A parallel plate capacitor with plate area \( A \) and plate separation \( d \) has a capacitance \( C_0 \). A slab of dielectric constant \( K \) having area \( A \) and thickness \( \left(\frac{d}{4}\right) \) is inserted in the capacitor, parallel to the plates. Find the new value of its capacitance.

Hint:

When a dielectric slab partially fills the gap in a capacitor, treat the system as two capacitors in series. The total capacitance is calculated using: \[ \frac{1}{C} = \frac{d_1}{K \varepsilon_0 A} + \frac{d_2}{\varepsilon_0 A} \] Always break it into dielectric and air segments before applying the formula.

Answer 1

Step 1: Understanding the arrangement: The total distance between the plates is \( d \), and a dielectric slab of thickness \( \frac{d}{4} \) and dielectric constant \( K \) is inserted between the plates. The remaining part of the gap is \( \frac{3d}{4} \), filled with air. Step 2: Concept used — Series Combination of Capacitors:
The capacitor is effectively divided into two parts in series: - One with dielectric \( K \) of thickness \( d_1 = \frac{d}{4} \) - One with air (dielectric constant 1) of thickness \( d_2 = \frac{3d}{4} \) The capacitance for each segment: \[ C_1 = \frac{K \varepsilon_0 A}{d_1} = \frac{4K \varepsilon_0 A}{d}, \quad C_2 = \frac{\varepsilon_0 A}{d_2} = \frac{4 \varepsilon_0 A}{3d} \] Step 3: Applying Series Formula:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{4K \varepsilon_0 A} + \frac{3d}{4 \varepsilon_0 A} \] \[ \frac{1}{C} = \frac{d}{\varepsilon_0 A} \left( \frac{1}{4K} + \frac{3}{4} \right) \Rightarrow C = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4K}} \]