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

#### Solution

Capacitors b and c are in parallel; their equivalent capacitance is 20 µF.

Thus, the net capacitance of the circuit is given by

`1/C_"net" = 1/10 + 1/20 + 1/10`

`⇒ 1/C_"net" = (2+1+2)/20 = 5/20`

`⇒ C_"net" = 4 "uF"`

The total charge of the battery is given by

`Q = C_"net"V = (4 "uF") xx (100 "V") = 4 xx 10^-4 C`

For a and d,

`q = 4 xx 10^-4 "C" and "C" = 10^-5 "F"`

`therefore E = q^2/(2C) = (4 xx 10^-4)/(2 xx 10^-5)`

`⇒ E = 8 xx 10^-3 "J" = 8 "mJ"`

For b and c,

`q = 4 xx 10^-4 "C" and "C"_(eq) = 2"C" = 2 xx 10^-5 "F"`

`therefore V = q/C_(eq) (4 xx 10^-4)/(2 xx 10^-5) = 20 V`

`⇒ E = 1/2 CV^2`

`⇒ E = 1/2 xx 10^-5 xx 400`

`⇒ E = 2 xx 10^-3 "J" = 2 "mJ"`