Question

The capacity of a parallel plate air capacitor is 10µF. As shown in the figure this capacitor is divided into two equal parts; these parts are filled by media of dielectric constant K₁ =2 and K₂ =4. Capacity of this arrangement will be :

• A. 20µF
• B. 30μF
• C. 10µF
• D. 40μF

Answer 1

Answer: (B)

30µF

Answer 2

The capacity of a parallel plate air capacitor is given by:

$C_0 = \frac{\epsilon_0 A}{d}$

Given $C_0 = 10 \mu F$.

As shown in the figure, the capacitor is divided into two equal parts along its area, which are then filled with dielectric media. This arrangement can be considered as two capacitors connected in parallel, as they share the same potential difference across their plates.

For the first part: Area $A_1 = A/2$ Distance $d_1 = d$ Dielectric constant $K_1 = 2$ The capacitance of this part, $C_1$, is:

$C_1 = K_1 \frac{\epsilon_0 (A/2)}{d} = K_1 \frac{1}{2} \left(\frac{\epsilon_0 A}{d}\right)$

Substituting $C_0 = \frac{\epsilon_0 A}{d}$:

$C_1 = \frac{K_1}{2} C_0$

For the second part: Area $A_2 = A/2$ Distance $d_2 = d$ Dielectric constant $K_2 = 4$ The capacitance of this part, $C_2$, is:

$C_2 = K_2 \frac{\epsilon_0 (A/2)}{d} = K_2 \frac{1}{2} \left(\frac{\epsilon_0 A}{d}\right)$

Substituting $C_0 = \frac{\epsilon_0 A}{d}$:

$C_2 = \frac{K_2}{2} C_0$

Since these two capacitors are in parallel, their equivalent capacitance $C_{eq}$ is the sum of their individual capacitances:

$C_{eq} = C_1 + C_2$ $C_{eq} = \frac{K_1}{2} C_0 + \frac{K_2}{2} C_0$ $C_{eq} = \frac{C_0}{2} (K_1 + K_2)$

Now, substitute the given values: $C_0 = 10 \mu F$ $K_1 = 2$ $K_2 = 4$

$C_{eq} = \frac{10 \mu F}{2} (2 + 4)$ $C_{eq} = 5 \mu F \times 6$ $C_{eq} = 30 \mu F$