Through resistanceless leads, a capacitor of capacitance 12.0 F is connected to a battery of emf 6.00 V and internal resistance 1.00 . Determine the rate at which the energy stored in the capacitor is increasing 12.0 s after the connections are made.

Step-by-step solution

The energy stored in a capacitor is given by U = q^2 2C . The rate at which the energy is increasing is dU dt = q C dq dt = V_c I . For an RC circuit during charging, the voltage across the capacitor and the current are given by: V_c = E(1 - e^ -t/RC ) I = E R e^ -t/RC The time constant of the circuit is = RC = (1.00 )(12.0 F ) = 12.0 s . At the given time t = 12.0 s , we have t = . Substituting this into the equations: V_c = E(1 - e⁻¹) I = E R e⁻¹ The rate of increase of energy is: dU dt = E(1 - e⁻¹) E R e⁻¹ = E^2