A capacitor of capacitance `100(mu)F` is connected across a battery of emf6.0V through a resistance of `20k(Omega)`for 4.0s after the battery is disconnecting?

Explanation

`c = 100 muF, E mu f = 6V, `
` r = 20 K , T = 4s `
Charging ,
`Q = CV (1-e^(-t/RC)) `
` = 6 xx (10^-4) (1 - e^-2) `
` = 5.187 xx (10^-4)C `
Discharging
` Q = q (e^(-t/RC))`
` = 5.184 xx (10^-4) xx e^-2 `
` = 0.7 xx (10^4)C = 70 mu C ` .