A`100 p F` capacitor is charged o a potential difference of` 24 V`. It is connected to an uncharged capacitor of capaciance` 20 p F `What will be the new polential diffeernce across the 100 pF capacitor?

Explanation

`C_1 = 100 pF , V = 25 V `
Uncharged capacitor of capacitance
`C2 = 20 pF `
Total charge `q = C_1 V = 24 xx 100 pF`
When the charge capacitor is connected with the uncharged capacitor
` q_1 + q_2 = 24 xx 100 qF and `q_1 / C_1 = q_1 / C_1`
q_1 / 100 = q_1 / 20 `
or q_1 = 5 q_2`
From equation (i) and (ii) , we have
` q_1 = 20 xx 100pF`
`v_1 = q_1 / C_1 `
`= 20 xx 100 pF / 20 pF = 20 V`