Each capacitor in figure has a capacitance of `10muF `. The emf of the battery is `100V` . Find the energy stored in each the four capacitors.

Explanation

`C = 10 mu F = 10 xx 10^(-6) F`
For a and d ,`q= 4xx 10^(4) C, C = 10^(-5)F. So ` E = (1) /(2) xx 10^(-5) xx 40 xx 40`
` = 8 xx 10^(-3) J = 8 m J `
for b and c
` q = 4 xx 10^(-4) C` ,
`C_eq = 2C = 2 xx 10^(-5)F`,
`v = 20 V`
So, ` E = 1 / 2 Cv^(2) `
`= 1 / 2 xx 10^(-5) xx 400 `
` = 2 xx 10^(-3) J = 2m J`.