A `5.0muF` capacitor is charged to `12V`. The positive plate of this capacitor is now connected to the negative terminal of a `12V`bettery and vide versa . Calculate the head developed in the connecting wires.

Explanation

When the capacitor is connected to the battery , a charge ` Q = CE` appears on one plate and - on the other. When the polarity is reversed , a charge - Q appears on the 1st plate + Q on the 2nd . A charge 2Q, therefore passes through the battery from the - ve to the + ve teeminal .
The battery does a work.
` W = Q xx E = 2 QE = 2 CE^(2)`
In this process , the energy stored in the capacitor is the same in the two cases. Thus the work done by the battery apperas as heat in the connecting wires . Heat produces is therefore,
` 2CE^(2) = 2 xx 5 xx 10^(-6) xx 144 `
`[have C = 5 mu F, v = E = 12V] `
` = 144 xx 10^(-5) J = 1.44 mJ`,