47-A capacitor of capacitor `5.00 muF` is charged to` 24.0 V` and another capacitor of capacitance `6.0muF` is charged to `12.0V` (a) Find the energy stored in each capacitor .(b)The positive plates of the first capcitor is now connnected to the negative plate of the second and vice versa . Find the new charges on the capacitors .(c)Find the loss of electrostatic energy during the process .(d)Where does this energy go?

Explanation

(a) `C_1 = 5 mu F, V_1 24 V `
q_1 = C_1 v_1 = 25 xx 24 = 120 mu C`
and ` C_2 = 6 mu F , V_2 = R `
`q_2 = C_2 v_2 = 6 xx 12 = 72 mu C`
Energy stored on first capacitor
` E_1 = 1 / 2 C_1 V_1^(2)`
`= 1440 J = 1.44 mJ`
Energy stored on second capacitor
` E_2 = 1 / 2 C_1 V_2^(2) `
` 432 J = 0.432 J `
(b ) `C_1 V_1 = C_2 V_2`
Let the effective potential = V
` V = 4.36`
The new charge
`C_1 V = 5 xx 4.36 = 21.8 mu C`
and C_2 V = 6 xx 4.36 = 26.16 `
`= 26.2 mu C`
(c) ` U_1 = 1 /2 C_1 V^(2) `
and U_2 = 1 / 2 C__2 V^(2) `
` U_f = 1 /2 V^(2) (C_1 + C_2) `
`= 1 /2 (4.36)^(2) (5+6)`
` = 1 / 2 xx 19 xx 11 `
` 104 . 5 xx 10^(-6) J `
`= 0.1045 mJ`
But, `U_i = 1.44 + 0.433 = 1.873`
The loss in KE `= 1.873 - .1045 `
`= 1.7687 = 1.77 mJ `
(d) The energy is destroyed as heat .