#### Question

A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20 pF. What will be the new potential difference across the 100 pF capacitor?

#### Solution

Given :

`C_1 = 100 "pF"`

V = 24 V

Charge on the capacitor `q = C_1V = 24 xx 100 "pC"`

Capacitance of the uncharged capacitor, `C_2 = 20 "pF"`

When the charged capacitor is connected with the uncharged capacitor, the net charge on the system of the capacitors becomes

`q_1 + q_2 = 24 xx 100 "qC"` ....(i)

The potential difference across the plates of the capacitors will be the same.

Thus,

`q_1/C_1 = q_2/C_2`

⇒ `q_1/100 = q_2/20`

⇒ `q_1 = 5q_2` ....................(ii)

From eqs. (i) and (ii), we get

`q_1+q_1/5 = 24 xx 100 "pc"`

⇒ `6q_1 = 5 xx 24 xx 100 "pC"`

⇒ `q_1 = (5xx24xx100)/(6) "pC"`

Now ,

`V_1 = q_1/C_1`

⇒ `(5xx24xx100 "pC")/(6xx100 "pF")` = 20 V