A capacitor with capacitance 5,mu F is charge | vedclass

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Question

(A) The charge $q$ on the capacitor remains constant because the capacitor is isolated after charging.Initial capacitance $C_i = 5\,\mu F = 5 	imes 1

Vedclass · Accepted Answer

$3.75	imes 10^{-6}\,J$

Vedclass · Answer

(A) The charge $q$ on the capacitor remains constant because the capacitor is isolated after charging.Initial capacitance $C_i = 5\,\mu F = 5 	imes 10^{-6}\,F$.Final capacitance $C_f = 2\,\mu F = 2 	imes 10^{-6}\,F$.Charge $q = 5\,\mu C = 5 	imes 10^{-6}\,C$.The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Work done $W = \Delta U = U_f - U_i = \frac{q^2}{2C_f} - \frac{q^2}{2C_i} = \frac{q^2}{2} \left( \frac{1}{C_f} - \frac{1}{C_i} ight)$.Substituting the values:$W = \frac{(5 	imes 10^{-6})^2}{2} \left( \frac{1}{2 	imes 10^{-6}} - \frac{1}{5 	imes 10^{-6}} ight)$.$W = \frac{25 	imes 10^{-12}}{2} \left( \frac{5 - 2}{10 	imes 10^{-6}} ight) = \frac{25 	imes 10^{-12}}{2} \left( \frac{3}{10 	imes 10^{-6}} ight)$.$W = \frac{75 	imes 10^{-12}}{20 	imes 10^{-6}} = 3.75 	imes 10^{-6}\,J$.

$A$ capacitor with capacitance $5\,\mu F$ is charged to $5\,\mu C.$ If the plates are pulled apart to reduce the capacitance to $2\,\mu F,$ how much work is done?

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