Question

Sol. [Key Concept: Model the half-filled gap as two capacitors in series; percentage increase $= (C_{f} - C)/C \times 100$.]

Initial capacitance

$$C_{i} = \frac{A \in_{0}}{\text{ }d} = C$$

Final capacitance

$$\begin{matrix} & C_{1} = \frac{A \in_{0} \times 5}{\frac{d}{2}} = 10C \\ & C_{2} = \frac{A \in_{0}}{\frac{d}{2}} = 2C \\ & C_{f} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} = \frac{10 \times 2}{10 + 2}C = \frac{5}{3}C \\ & \%\text{~}\text{increase}\text{~} = \frac{\frac{5}{3}C - C}{C} \times 100 = 66.67 \end{matrix}$$