Question

An energy storage configuration of two large square metallic parallel plates in between terminals A and B utilizing air ($K_{air}$ = 1) as insulation barrier exhibits capacitance C. An insulating slab with dielectric constant K matching the inter-surface spacing partially occupies one-fourth of the geometric volume as depicted. Determine the new capacitance

• A. (K+3)C/4
• B. (K+2)C/4
• C. (K+1)C/4
• D. KC/4

Answer 1

Answer: (A)

(K+3)C/4

Answer 2

The initial capacitance of the parallel plate capacitor with air ($K_{air}=1$) is given by $C = \frac{\epsilon_0 A}{d}$, where $A$ is the area of the plates and $d$ is the separation.

The dielectric slab with dielectric constant $K$ occupies one-fourth of the geometric volume. This means the dielectric covers one-fourth of the area, and the remaining three-fourths is filled with air. This configuration can be treated as two capacitors in parallel: one with dielectric and one with air.

Let the total area of the plates be $A$. The area covered by the dielectric slab is $A_1 = A/4$. The dielectric constant is $K_1 = K$. The area covered by air is $A_2 = 3A/4$. The dielectric constant is $K_2 = K_{air} = 1$. Both parts have the same separation $d$.

The capacitance of the part with the dielectric is: $C_1 = \frac{K_1 \epsilon_0 A_1}{d} = \frac{K \epsilon_0 (A/4)}{d} = \frac{K}{4} \frac{\epsilon_0 A}{d}$

The capacitance of the part with air is: $C_2 = \frac{K_2 \epsilon_0 A_2}{d} = \frac{1 \cdot \epsilon_0 (3A/4)}{d} = \frac{3}{4} \frac{\epsilon_0 A}{d}$

Since these two parts are in parallel, the new total capacitance $C_{new}$ is the sum of $C_1$ and $C_2$: $C_{new} = C_1 + C_2 = \frac{K}{4} \frac{\epsilon_0 A}{d} + \frac{3}{4} \frac{\epsilon_0 A}{d}$ $C_{new} = \frac{\epsilon_0 A}{4d} (K + 3)$

We know that the original capacitance $C = \frac{\epsilon_0 A}{d}$. Substituting this into the expression for $C_{new}$: $C_{new} = \frac{K+3}{4} C$

This matches option (1).