At `t=0` switch `S` is closed .
(a) After how much time , charge on capacitor is 50% of maximum charge stored on capacitor.
(b) After how much time, energy stored in capacitor is halt of maximum energy stored in capacitor.

Explanation

(a) `q = CE (1-e^(-t//RC))`
`q_(max) = CE, q = (q_(max))/(2) = 1/2 CE`
`1/1 CE = CE(1-e^(-t//RC))`
`e^(-t//RC)=2`
`t/(RC)`log_(e) e = log_(e) 2 = 0.693RC`
(b) `U =(q^2)/(2C) = 1/2 CE^(2)(1-e^(-t//RC))^(2)`
`U_(max) = 1/2 CE^2, U = (U_(max))/(2) = (CE^2)/(4)`
`(CE^2)/(4) = (CE^2)/() (1-e^(-t//RC))^(2)`
`(1-e^(-t//RC))^(2) = 1/2 implies (1-e^(-t//RC))=1/(sqrt(2))`
`e^(-t//RC)=1-1/(sqrt(2)) = (sqrt(2)-1)/(sqrt(2))`
`e^(-t//RC) = (sqrt(2))/(sqrt(2)-1) = (sqrt(2))/(sqrt(2)-1)* ((sqrt(2)-1))/((sqrt(2)+1)) = (2+sqrt(2))`
`t/(RC ) log_(e) e = log_(e) (2+sqrt(2))` ltbr .`t = RC log_(e) (2+sqrt(2))`.