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Question

Physics Question on electrostatic potential and capacitance

A capacitor is charged with a battery and energy stored is UU. After disconnecting battery another capacitor of same capacity is connected in parallel with it. Then energy stored in each capacitor is:

A

U2\frac U2

B

U4\frac U4

C

4U4U

D

2U2U

Answer

U4\frac U4

Explanation

Solution

Parallel combination of Capacitor

Energy stored (U)=q22CEnergy\ stored\ (U) =\frac {q^2}{2C}

After connecting with another capacitor

Vcommon=q1+q2C1+C2=q+0C+C=q2CV_{common}=\frac {q_1+q_2}{C_1+C_2}=\frac {q+0}{C+C}=\frac {q}{2C}

∴Energy on each capacitor =12CVcommon2\frac {1}{2} CV^2_{common} = 12(q2C)2\frac 12 (\frac {q}{2C})^2 = U4\frac U4

So, the correct option is (B): U4\frac U4