A 10.0 mF capacitor that is initially uncharged is connected in series with a 10.0 M$\Omega$ resistor and an emf source with E = 100 V. Just after the connection is made, the rate of increase of energy in the capacitor is
Answer & explanation
Correct answer: option 2
$ U = \frac{Q^2}{2C}$
$ \frac{dU}{dt} = \frac{Q}{C} \frac{dQ}{dt} = \frac{Q_0 e^{\frac{-t}{RC}}}{C} \frac{Q_0}{RC} e^{{-t}{RC}}$
$ \frac{dU}{dt} = \frac{Q_0^2}{RC^2} e^{\frac{-2t}{RC}}$
At t= 0 , $\frac{dU}{dt} = \frac{Q_0^2}{RC^2} = \frac{E^2}{R} = 10^{-3} J/s$