Question

There is a parallel plate capacitor of capacitance $C$. If half of the space is filled with dielectric of dielectric constant $k = 5$ as in the figure. Find percentage increase in capacitance.

Answer 1

Step 1: Identify initial capacitance $C_i$.
$C_i = C = \frac{\epsilon_0 A}{d}$.

Step 2: Calculate the equivalent capacitance $C_f$ after filling half the gap.
This setup acts as two capacitors in series: one with air ($d_1 = d/2$) and one with dielectric ($d_2 = d/2, k=5$).
$C_1 = \frac{\epsilon_0 A}{d/2} = 2C$.
$C_2 = \frac{k \epsilon_0 A}{d/2} = 10C$.

Step 3: Find series combination $C_f$.
$C_f = \frac{C_1 C_2}{C_1 + C_2} = \frac{2C \times 10C}{2C + 10C} = \frac{20C^2}{12C} = \frac{5}{3}C$.

Step 4: Calculate percentage increase.
Percentage Increase = $\frac{C_f - C_i}{C_i} \times 100 = \frac{\frac{5}{3}C - C}{C} \times 100 = \frac{2}{3} \times 100 = 66.67\%$.

Final Answer: 66.67%.