Electrostatic Potential and Capacitance Part 2

Concept:The capacitance of a parallel plate capacitor depends on the area of plates, separation, and the dielectric constant.This arrangement is broken into individual capacitors and then combined using series and parallel rules.Explanation:Let the original capacitance without dielectric be $C = \frac{\epsilon_0 A}{d}$.The gap $d$ is divided into two halves, each of thickness $\frac{d}{2}$.For the top half: dielectric $K_1 = 2$ covers the full area $A$.Its capacitance is $C_1 = \frac{K_1 \epsilon_0 A}{d/2} = \frac{2 \epsilon_0 A}{d/2} = 4C$.For the bottom half: two dielectrics $K_2 = 3$ and $K_3 = 5$ each cover area $\frac{A}{2}$ and thickness $\frac{d}{2}$.$C_2 = \frac{3 \epsilon_0 (A/2)}{d/2} = 3C$.$C_3 = \frac{5 \epsilon_0 (A/2)}{d/2} = 5C$.Since $C_2$ and $C_3$ share the same potential difference, they are in parallel: $C_{23} = C_2 + C_3 = 8C$.Now $C_1$ (top half) and $C_{23}$ (bottom half) are in series because the charge flows through one then the other.The equivalent capacitance: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_{23}} = \frac{1}{4C} + \frac{1}{8C} = \frac{3}{8C}$.Thus $C_{\text{eq}} = \frac{8C}{3}$.Given $C_{\text{eq}} = \frac{n}{3}C$, we get $\frac{8C}{3} = \frac{n}{3}C$, so $n = 8$.Answer:$n = 8$, which corresponds to option A.