A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (1/4)d, where d is the separation of the plates. If the slab is inserted between the plates of the capacitor, the potential difference between the plates will be
(Given: Potential difference across the capacitor without dielectric is Vo)
Answer & explanation
Correct answer: option 2
The correct answer is Option (1) → $V=V_0\left(\frac{3K+1}{4K}\right)$
Given:
Dielectric constant of slab, $K$
Thickness of slab, $t = \frac{d}{4}$
Plate separation, $d$
Potential difference without dielectric, $V_0$
The capacitor can be considered as two capacitors in series:
- One with dielectric: $C_1 = \frac{K \epsilon_0 A}{t} = \frac{K \epsilon_0 A}{d/4} = \frac{4 K \epsilon_0 A}{d}$
- One without dielectric: $C_2 = \frac{\epsilon_0 A}{d - t} = \frac{\epsilon_0 A}{3d/4} = \frac{4 \epsilon_0 A}{3 d}$
Total capacitance for series combination:
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{(4 K \epsilon_0 A / d)} + \frac{1}{(4 \epsilon_0 A / 3 d)} = \frac{d}{4 K \epsilon_0 A} + \frac{3 d}{4 \epsilon_0 A} = \frac{3K + 1}{4 K} \cdot \frac{d}{\epsilon_0 A}$
So, $C = \frac{4 K \epsilon_0 A}{(3K + 1)d}$
Potential difference across capacitor: $V = \frac{Q}{C}$. Let $Q$ be the same charge as initially, then:
$V = Q/C = Q \cdot \frac{(3K + 1)d}{4 K \epsilon_0 A} = \frac{3K + 1}{4K} V_0$
Potential difference across the plates: $V = \frac{3K + 1}{4K} V_0$