A capacitor having a capacitance of `100muF` is chargeed to a potential difference of 24V . The charging bettery is disonnected and the capacitor is connected to another bettery of emf `12V` with the positive plate of the capacitor joined with the positive terminal of the bettery . (a ) Find the charges on the capacitor before and after the reconnection . (b ) Find the charge flown through the 12V bettery . (c ) Is work done by the bettery or is it done on the bettery ? find its magnitude . (d ) Find the decrease is electrostatic field energy . (e ) Find the best developed during the flow of charge after reconnection.

Explanation

(a) Before reconnection
`C = 100 mu F , V = 24 V `
So, ` q = Cv = 2400 mu F `
After reconnection when
` C = 100 mu F , V = 12 V `
So, ` q = Cv = 1200 mu C `
(b) `C = 100 , q = 12 V `
So, ` q = Cv = 100 xx 12 = 1200 mu C ` (c ) We know w = vq = 12 xx 1200 `
` = 14400 J = 14.4 mJ ` The work done on the battery (d ) Initial electrostaic field energy U_1
` = 1 / 2 Cv_1 ^(2)`
Final eletrostaic field energy U_f
` = 1 / 2 Cv_2^(2) `
Decrease in electrostasic field energy ` = 1 /2 (Cv_1^(2) - Cv_2^(2))`
`= 1 / 2 C (v_1^(2) - v_2^(2))`
` = 1 /2 = 100 [(24)^(2) - (12)^(2)]`
`= 1 / 2 xx 100 xx (576 - 144 ) `
`= 21600 J`
Energy `= 21600 J = 21.6 mJ`
(e) After reconnection
` C = 100 mu F , v = 12 V`
The energy appeared
` = 1 / 2 Cv^(2) = 100 xx 144`
` = 7200 J = 7.2 mJ`
This amount of energy is devloped as heat when the charge flows through the capacitor .