A parallel-plate capacitor has plate area `20cm^(2)`,plate separation 1.0mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates.This capacitor is joined to a battery of emf 6.0V through a `100k(Omega)`resistor.find the energy of the capacitor `8.9(mu)s`after the connections are made.

Explanation

`A = 20 cm^2, d = 1 mm , K = 5, `
` t = 8.9 xx (1^-6)C epsilon = 6V `
` R = 100 xx (10^-3) Omega `
` C = ((K epsilon_0 A)/d)`
` = (5 xx 8.85 xx (10^-12) xx 20 xx (10^-4)/1 xx (10^-3))`
` = (10 xx 8.85 xx (10^-12) xx 20 xx (10^-4)/ 1 xx (10^-3))`
` = 88.5 xx (10^-12)`
` Q = EC (1- e^(-t/rc))`
` = 6 xx 88.5 xx (10^-12)(1-e)`
` = 530.67 xx (10^-12) C `
` Energy = 1/2 xx ((500 xx 530)/ (88 xx (10^12)))`
` = ((500 xx 530.97)/ (88 xx 2)) xx (10^-12)` .