Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination.

Explanation

Here `C_1 = C_2 = 12 pF = 12 xx 10^(-12) F and V = 50 V`.
When the two capacitors are connected in series, the equivalent capacitance
`C_s = (C_1C_2)/(C_1 +C_2) = ((12 xx 10^(-12))xx(12 xx 10^(-12)))/((12 xx 10^(-12) + 12 xx 10^(-12)))=6 xx 10^(-12) F`
and energy stored in the series combination
`u_s = 1/2 C_s V^2 = 1/2 xx 6 xx 10^(-12) xx (50)^2 = 7.5 xx 10^(-9)J = 7.5 nJ`
When the same capacitors are connected in parallel, the equivalent capacitance
`C_p =C_1 + C_2 = 12 xx 10^(-12) + 12 xx 10^(-12) = 24 xx 10^(12)F`
`:.` Energy stored in parallel combination
`u_p =1/2 C_p V^2 = 1/2 xx 24 xx 10^(-12) xx (50)^2 = 30 xx 10^(-9)J = 30 nJ`