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(C) The original capacitance is given by $C = \frac{A \varepsilon_0}{d} = 4 \, \mu F$.When the space is filled as shown,the system acts like two capac

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$6$

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(C) The original capacitance is given by $C = \frac{A \varepsilon_0}{d} = 4 \, \mu F$.When the space is filled as shown,the system acts like two capacitors in series,each with plate separation $d/2$.The first capacitor (air-filled) has capacitance $C_1 = \frac{A \varepsilon_0}{d/2} = 2 \left( \frac{A \varepsilon_0}{d} ight) = 2C$.The second capacitor (dielectric-filled) has capacitance $C_2 = \frac{K A \varepsilon_0}{d/2} = 2K \left( \frac{A \varepsilon_0}{d} ight) = 2KC = 2(3)C = 6C$.Since they are in series,the equivalent capacitance $C_{	ext{new}}$ is:$C_{	ext{new}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(2C)(6C)}{2C + 6C} = \frac{12C^2}{8C} = 1.5C$.Substituting $C = 4 \, \mu F$:$C_{	ext{new}} = 1.5 	imes 4 \, \mu F = 6 \, \mu F$.

$A$ parallel plate capacitor with plate area $A$ and plate separation $d$ has a capacitance of $4 \, \mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K = 3$ (as shown in figure) will be .........$ \mu F$.

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