A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system: Option 1) increases by a factor of 4 Option 2) decreases by a factor of 2 Option 3) remains the same Option 4) increases by a factor of 2

As we discussed in

Energy Stored -

$\dpi{100} U=\frac{1}{2}CV^{2}=\frac{1}{2}QV=\frac{Q^{2}}{2C}$

-

when the capacitor is charged by a battery of potential V, then energy stored in the capaciter

$V_{i}=\frac{1}{2}Cv^{2}$

When the battery is removed and another identical unchanged capacitor is connected in parallel

common potential

$V^{1}=\frac{Cv}{C+c}=\frac{V}{2}$

$V_{f}=\frac{1}{2}(2c)(\frac{v}{2})^{2}$

$\frac{1}{4}CV^{2}$

$V_{f}=\frac{V_{1}}{2}$

Option 1)

increases by a factor of 4

Option is incorrect

Option 2)

decreases by a factor of 2

Option is correct

Option 3)

remains the same

Option is incorrect

Option 4)

increases by a factor of 2

Option is incorrect

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