A parallel-plate capacitor having a plate area of 40 cm ^2 and a separation of 0.10 mm between the plates is connected to a battery with an emf of 2.0 V through a 16 resistor. Determine the electric field inside the capacitor 10 ns after the connections are established.

Step-by-step solution

Area of the plates, A = 40 cm ^2 = 4 10⁻³ m ^2 Separation between plates, d = 0.10 mm = 10⁻⁴ m Capacitance, C = ₀ A d = 8.85 10⁻¹² 4 10⁻³ 10⁻⁴ = 3.54 10⁻¹⁰ F Time constant of the RC circuit, = RC = 16 3.54 10⁻¹⁰ = 5.664 10⁻⁹ s = 5.664 ns The voltage across the capacitor at time t is given by: V(t) = V₀ (1 - e^ -t/ ) Substituting t = 10 ns and = 5.664 ns : V(10 ns ) = 2.0 (1 - e^ - 10 5.664 ) = 2.0 (1 - e^ -1.765 ) V(10 ns ) 2.0 (1 - 0.171) = 2.0 0.829 = 1.658 V The electric field inside the capacitor is: E = V d =