A 4 ;mu F capacitor is charged by a 200 ;V su | vedclass

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(A) Initial capacitance $C_1 = 4 \;\mu F = 4 	imes 10^{-6} \;F$ and supply voltage $V_1 = 200 \;V$.Initial electrostatic energy $E_1 = \frac{1}{2} C_

Vedclass · Accepted Answer

$2.67 	imes 10^{-2} \;J$

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(A) Initial capacitance $C_1 = 4 \;\mu F = 4 	imes 10^{-6} \;F$ and supply voltage $V_1 = 200 \;V$.Initial electrostatic energy $E_1 = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} 	imes 4 	imes 10^{-6} 	imes (200)^2 = 8 	imes 10^{-2} \;J$.When connected to an uncharged capacitor $C_2 = 2 \;\mu F$,the common potential $V$ is given by $V = \frac{C_1 V_1}{C_1 + C_2} = \frac{4 	imes 200}{4 + 2} = \frac{800}{6} = \frac{400}{3} \;V$.Final energy $E_2 = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} 	imes (6 	imes 10^{-6}) 	imes (\frac{400}{3})^2 = 3 	imes 10^{-6} 	imes \frac{160000}{9} = \frac{16}{3} 	imes 10^{-2} \approx 5.33 	imes 10^{-2} \;J$.Energy lost $\Delta E = E_1 - E_2 = 8 	imes 10^{-2} - 5.33 	imes 10^{-2} = 2.67 	imes 10^{-2} \;J$.

$A$ $4 \;\mu F$ capacitor is charged by a $200 \;V$ supply. It is then disconnected from the supply and connected to another uncharged $2 \;\mu F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

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