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A capacitor of capacitance 2 \mu \mathrm{f} is connected to a call of emf 20 \mathrm{~V}. The plates of a capacitor are drawn apart slowly to double the distance between them. The work done by the external agent on the plates.

Option: 1

200 \mu \mathrm{J}


Option: 2

4 \times 10^3 \mu \mathrm{J}


Option: 3

2 \times 10^3 \mathrm{\mu J}


Option: 4

100 \mu J


Answers (1)

best_answer

\mathrm{using, v_i+w_{\text {battery }}+w_{\text {agent }}=U_f+ loss\, \, As \, \, the \, \, plates\, \, are\, \, pulled \, \, \, slowly\, \, so,\, \, \, loss =0 }

\mathrm{\begin{aligned} & \text { Charge flow } a=Q_f-Q_i=1 \times 20-2 \times 20 \\ & =-20 \mu c \\ & \therefore \begin{aligned} \frac{1}{2} \times 2 \times(20)^2-20 \times 20 * \omega_{\text {agent }} & =\frac{1}{2} \times 1 \times 20^2+0 \\ 400-400+\omega_{\text {agent }} & =200 \\ W_{\text {agent }} & =200 \mu \mathrm{J} \end{aligned} \end{aligned} }

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rishi.raj

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