A parallel-plate capacitor of plate area `40 cm^2` and separation between the plates `0.10 mm` is connected to a battery of `emf 2.0 V` through a `16 Omega` resistor. Find the electric field in the capacitor `10 ns` after the connections are made.

Explanation

`A = 40 cm^2 = 40 xx (10^-4) (m^2) `
` D = (0.1) mm = 1 xx (10^-4) m, `
` R = 16 Omega , E_mf = 2 V `
` C = epsilon_0A/d`
` = (8.85 xx (10^-12) xx 40 xx (10^-4)/1 xx (10^-4))`
` = 35.4 xx (10^-11) F `
` Now, E = Q/AE_0 (1-e^(-t/rc))`
` = (35.4 xx (10^-11) xx 2 / 8.85 xx (10^-12) xx 40 xx (10^-4)) (1- e^(-1.76))`
` = 1.655xx (10^-4) = 1.7 xx (10^-4) V/m`