The capacitance of a parallel plate capacitor | vedclass

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(B) Given the initial capacitance $C = \frac{\varepsilon_0 A}{d} = 2.5 \mu F$.When the capacitor is half-filled with a dielectric as shown,it acts as

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$3$

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(B) Given the initial capacitance $C = \frac{\varepsilon_0 A}{d} = 2.5 \mu F$.When the capacitor is half-filled with a dielectric as shown,it acts as two capacitors in parallel,each with plate area $A/2$ and separation $d$.The first part is filled with air (dielectric constant $K_1 = 1$):$C_1 = \frac{\varepsilon_0 (A/2)}{d} = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} ight) = \frac{2.5 \mu F}{2} = 1.25 \mu F$.The second part is filled with a dielectric (dielectric constant $K_2 = K$):$C_2 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K}{2} \left( \frac{\varepsilon_0 A}{d} ight) = K 	imes 1.25 \mu F$.Since they are in parallel,the equivalent capacitance $C_{eq} = C_1 + C_2$.Given $C_{eq} = 5 \mu F$:$5 = 1.25 + 1.25 K$$3.75 = 1.25 K$$K = \frac{3.75}{1.25} = 3$.Therefore,the dielectric constant is $3$.

The capacitance of a parallel plate capacitor is $2.5 \mu F$. When it is half filled with a dielectric as shown in the figure,its capacitance becomes $5 \mu F$. The dielectric constant of the dielectric is

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