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# 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? - Physics

ConceptElectric Potential Difference

#### 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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Solution 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? Concept: Electric Potential Difference.
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