Question

A capacitor has capacitance \(C_0\) when there is no dielectric between its plates. Two slabs of dielectric constants \(K_1\) and \(K_2\), each having area equal to the area of the plates but thickness equal to half of the separation between the plates, are placed between the plates. Then the new capacitance is:

• A. \(C_0(K_1+K_2)\)
• B. \(C_0\left(\dfrac{K_1K_2}{K_1+K_2}\right)\)
• C. \(C_0\left(\dfrac{K_1+K_2}{K_1K_2}\right)\)
• D. \(2C_0\left(\dfrac{K_1K_2}{K_1+K_2}\right)\)

Hint:

When dielectric slabs are placed one after another along the electric field direction, they behave like capacitors connected in series. Always find the individual capacitances first and then apply the series combination formula.

Answer 1

The Correct Option is D

Step 1: Identify that the slabs are in series.
Each slab occupies half the gap (thickness d/2) and has the same plate area. They act as two capacitors in series: \( C_1 = \frac{2K_1\varepsilon_0 A}{d} = 2K_1 C_0 \) and \( C_2 = 2K_2 C_0 \).

Step 2: Find the equivalent series capacitance.
\[ C = \frac{C_1 C_2}{C_1+C_2} = \frac{2K_1 C_0\cdot2K_2 C_0}{2K_1 C_0+2K_2 C_0} = \frac{2K_1 K_2}{K_1+K_2}C_0 \] \[ \boxed{2C_0\!\left(\dfrac{K_1K_2}{K_1+K_2}\right)} \]