A metal plate of thickness 2 ,mm and area 36 | vedclass

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(C) When a metal plate of thickness $t = 2 \,mm$ is inserted into a parallel plate capacitor with plate spacing $d = 6 \,mm$,the air gaps on either si

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$25$

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(C) When a metal plate of thickness $t = 2 \,mm$ is inserted into a parallel plate capacitor with plate spacing $d = 6 \,mm$,the air gaps on either side of the metal plate are $d_1 = 3 \,mm$ and $d_2 = d - d_1 - t = 6 - 3 - 2 = 1 \,mm$.This arrangement is equivalent to two capacitors in series,each with area $A = 36 \pi \,cm^2 = 36 \pi 	imes 10^{-4} \,m^2$.The equivalent capacitance $C_{	ext{eq}}$ is given by $\frac{1}{C_{	ext{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d_1}{\varepsilon_0 A} + \frac{d_2}{\varepsilon_0 A} = \frac{d_1 + d_2}{\varepsilon_0 A}$.Substituting the values: $d_1 + d_2 = 3 \,mm + 1 \,mm = 4 \,mm = 4 	imes 10^{-3} \,m$.$C_{	ext{eq}} = \frac{\varepsilon_0 A}{d_1 + d_2} = \frac{1}{4 \pi 	imes 9 	imes 10^9} 	imes \frac{36 \pi 	imes 10^{-4}}{4 	imes 10^{-3}}$.$C_{	ext{eq}} = \frac{1}{36 	imes 10^9} 	imes \frac{36 	imes 10^{-4}}{4 	imes 10^{-3}} = \frac{10^{-4}}{4 	imes 10^{-3} 	imes 10^9} = \frac{1}{4} 	imes 10^{-10} \,F = 0.25 	imes 10^{-10} \,F = 25 	imes 10^{-12} \,F = 25 \,pF$.

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