The circular plates A and B of a parallel plate air capacitor have a diameter of 0.1 m and are `2 xx 10^(-3) m` apart. The plates C and D of a similar capacitor have a diameter of `0.12 m` and are `3 xx 10^(-3)m` apart. Plate A is earthed. Plates B and D are connected together. Plate C is connected to the positive pole of a 120V battery whose negative is earthed. The energy stored in the system is

Options

A. `0.1224 mu J`
B. `0.2224 muJ`
C. `0.3224 muJ`
D. `0.4224 muJ`

Answer

A

Explanation

As shown in figure, according to given problem the two capacitors `C_(1)` and `C_(2)` are in series

`:. C_(1) = (epsilon_(0)A_(1))/(d_(1)) = (epsilon_(0) xx pi (0.05)^(2))/(2 xx 10^(-3)) = (5)/(4) pi epsilon_(0)`
and ` C_(2) = (epsilon_(0)A_(2))/(d_(2)) = (epsilon_(0) xx pi (0.06)^(2))/(3 xx 10^(-3)) = (6)/(5) pi epsilon_(0)`
The capacity of the system is given by
`C = (C_(1)C_(2))/(C_(1) + C_(2)) = (pi epsilon_(0)(5//4) xx (6//5))/((5//4) + (6//5)) = (30)/(49) pi epsilon_(0)`
or `C = (15)/(98) xx 4 pi epsilon_(0) = (15)/(98) xx (1)/(9 xx 10^(9)) = 17 xx 10^(-12)F = 17 pF`
And hence the energy stored in the system,
`U = (1)/(2)CV^(2) = (1)/(2) = (17 xx 10^(-12))(120)^(2) = 0.1224 mu J`