A parallel plate capacitor 'A' of capacitance `1 muF` is charged to a potential drop 100 volt and then disconnected form the battery. Now the capacitor 'A' is completely filled with a dielectric slap of dielectric constant `k = 4`. Another parallel plate capacitor 'B' of capacitance `2 muF` is charged to a potential drop `20` volt and then disconnected from the battery. Now the two capacitors 'A' and 'B' are connected with each other with opposite polarities. Then the heat dissipated after connecting the capacitors is

Options

A. `1.15 xx 10^(-3) J`
B. `1.25 xx 10^(-3) J`
C. `1.35 xx 10^(-3) J`
D. `1.45 xx 10^(-13) J`

Answer

C

Explanation

`Q_(A) = 1 xx 100 = 100 = 100muC`
`Q_(B) = 2 xx 20 = 40 muC`
When the two capacitors are connected with oopposite polarities
`q_(A) = 40muC, q_(B) = 20muC`
The heat dissipated after two capacitors connected
`DeltaH = U_(i) - U_(f) = [((100)^(2))/(2 xx 4) + ((40)^(2))/(2 xx 2)] - ((60)^(2))/(2 xx 6)`
`= (1250 + 400) - 300 = 1350 muJ = 1.35 xx 10^(-3) J`