A parallel plate air capacitor has a capacitance C. When it is half filled as show in figure with a dielectric constant K = 5, the percentage…

Initial capacitance of the air capacitor is given by:

$$C = \dfrac{\epsilon_0 A}{d}$$C=dϵ0​A​

When the capacitor is half-filled with a dielectric of thickness $d/2$d/2, it can be considered as two capacitors connected in series.

Capacitance of the dielectric-filled part:

$$C_1 = \dfrac{K \epsilon_0 A}{d/2} = \dfrac{2K \epsilon_0 A}{d} = 2KC$$C1​=d/2Kϵ0​A​=d2Kϵ0​A​=2KC

Capacitance of the air-filled part:

$$C_2 = \dfrac{\epsilon_0 A}{d/2} = \dfrac{2 \epsilon_0 A}{d} = 2C$$C2​=d/2ϵ0​A​=d2ϵ0​A​=2C

The equivalent capacitance $C_{eq}$Ceq​ of the series combination is:

$$C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{(2KC)(2C)}{2KC + 2C} = \dfrac{2K}{K+1} C$$Ceq​=C1​+C2​C1​C2​​=2KC+2C(2KC)(2C)​=K+12K​C

Substituting $K = 5$K=5:

$$C_{eq} = \dfrac{2(5)}{5+1} C = \dfrac{10}{6} C = \dfrac{5}{3} C$$Ceq​=5+12(5)​C=610​C=35​C

The increase in capacitance is:

$$\Delta C = C_{eq} - C = \dfrac{5}{3} C - C = \dfrac{2}{3} C$$ΔC=Ceq​−C=35​C−C=32​C

The percentage increase in capacitance is:

$$\% \text{ increase} = \dfrac{\Delta C}{C} \times 100 = \dfrac{2}{3} \times 100 \approx 66.67\%$$% increase=CΔC​×100=32​×100≈66.67%

Answer: $66.67$66.67