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

According to the problem, the initial capacitance of the parallel plate capacitor is \( C \). After half of the space is filled with a dielectric of dielectric constant \( k = 5 \), the capacitance changes. To calculate the percentage increase in capacitance, follow these steps:

Step 1: Initial Capacitance
The initial capacitance \( C_i \) of the parallel plate capacitor is given by: \[ C_i = \frac{\epsilon_0 A}{d}, \] where \( A \) is the area of the plates and \( d \) is the separation between them.

Step 2: Capacitance after filling half the space
The setup now consists of two capacitors in series: - One part with air (\( d_1 = d/2 \)) and capacitance \( C_1 = \frac{\epsilon_0 A}{d/2} = 2C \), - The other part with dielectric material (\( d_2 = d/2 \), \( k = 5 \)) and capacitance \( C_2 = \frac{k \epsilon_0 A}{d/2} = 10C \).

Step 3: Total Capacitance
Since the capacitors are in series, the total capacitance \( C_f \) is: \[ C_f = \frac{C_1 C_2}{C_1 + C_2} = \frac{2C \times 10C}{2C + 10C} = \frac{20C^2}{12C} = \frac{5}{3}C. \]

Step 4: Percentage Increase in Capacitance
The percentage increase in capacitance is: \[ \text{Percentage Increase} = \frac{C_f - C_i}{C_i} \times 100 = \frac{\frac{5}{3}C - C}{C} \times 100 = \frac{2}{3} \times 100 = 66.67\%. \]

Final Answer: The percentage increase in capacitance is \( 66.67\% \).