Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination ? If these were connected in parallel across the same battery, how much energy will be stored in the combination now? Also find the charge drawn from the battery in each case.

Explanation

For two identical capacitors connected in series, electrostatic energy stored,
`U_s = (1)/(2)C_sV^2`
`C_s = (C_1C_2)/(C_1+C_2)= (12xx12)/(12+12)= 6pF ,V = 50V`
`U_s= (1)/(2)xx6xx10^(-12)xx(50)^2 = 7.3 nJ`
For two identical capacitors connected in parallel, electrostatic energy stored, `U_p = (1)/(2)C_pV^2`
`C_p= C_1+C_2 = 12pF+12pF = 24xx10^(-12)F`
`U_p = (1)/(2)xx24xx10^(-12)xx(50)^2 = 30 nJ`
Charge drawn from the battery for series combination of two identical capacitors,
`Q_s = C_sV = 6xx10^(-12)xx50 = 300 pC`
Charge drawn from the battery for parallel combination of two capacitors,
`Q_p= C_pV = 24xx10^(-12)xx50= 1200pC`.