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(B) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$.When the capacitor is half-filled with a dielectric of dielectric constant $K=5$ in series (as

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$66.6$

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(B) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$.When the capacitor is half-filled with a dielectric of dielectric constant $K=5$ in series (as shown in the figure),the system acts as two capacitors in series,each with plate separation $d/2$.Capacitance of the dielectric part: $C_1 = \frac{K \varepsilon_0 A}{d/2} = \frac{2 K \varepsilon_0 A}{d} = \frac{10 \varepsilon_0 A}{d}$.Capacitance of the air part: $C_2 = \frac{\varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$.The equivalent capacitance $C'$ is given by:$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{10 \varepsilon_0 A} + \frac{d}{2 \varepsilon_0 A} = \frac{d}{\varepsilon_0 A} \left( \frac{1}{10} + \frac{1}{2} ight) = \frac{d}{\varepsilon_0 A} \left( \frac{1+5}{10} ight) = \frac{6d}{10 \varepsilon_0 A} = \frac{3d}{5 \varepsilon_0 A}$.Therefore,$C' = \frac{5}{3} \frac{\varepsilon_0 A}{d} = \frac{5}{3} C$.Percentage increase in capacitance = $\frac{C' - C}{C} 	imes 100 = \left( \frac{5/3 C - C}{C} ight) 	imes 100 = \left( \frac{5}{3} - 1 ight) 	imes 100 = \frac{2}{3} 	imes 100 = 66.67\% \approx 66.6\%$.

$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $5$,the percentage increase in the capacitance will be.....$\%$

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