Through a resistance of 100 , a 20 F capacitor is connected to a battery with an emf of 6.0 V . Determine the charge on the capacitor 2.0 ms after making the connections.

Step-by-step solution

The maximum charge on the capacitor is given by q₀ = CV . q₀ = 20 10⁻⁶ 6.0 = 120 C The time constant of the RC circuit is = RC . = 100 20 10⁻⁶ = 2.0 10⁻³ s = 2.0 ms The charge on the capacitor at any time t during charging is given by q(t) = q₀(1 - e^ -t/ ) . Substituting t = 2.0 ms and = 2.0 ms : q = 120(1 - e^ -2.0/2.0 ) = 120(1 - e⁻¹) Using e 2.718 , we have e⁻¹ 0.368 . q = 120(1 - 0.368) = 120 0.632 = 75.84 C Rounding to the nearest integer, the charge is approximately 76 C .