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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
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