A) \[\frac{1}{2}C{{V}^{2}}\]
B) \[\frac{3}{2}C{{V}^{2}}\]
C) \[-\frac{3}{2}C{{V}^{2}}\]
D) \[C{{V}^{2}}\]
Correct Answer: A
Solution :
Capacitance of capacitor \[C=\frac{{{\varepsilon }_{0}}A}{d}\] Initial energy stored in the capacitor \[{{U}_{i}}=\frac{1}{2}C{{V}^{2}}\] When the separation between the plates is doubled, its capacitance becomes \[C'=\frac{{{\varepsilon }_{0}}A}{2d}=\frac{1}{2}\frac{{{\varepsilon }_{0}}A}{d}=\frac{C}{2}\] As the battary is disconnected, so charged capacitor becomes isolated and charge on it will remain constant \[O'=Q\] \[C'V'=CV\] \[V'=\left( \frac{C}{C'} \right)V=\frac{C}{\frac{C}{2}}=2V\] Final energy stored in the capacitor \[{{U}_{f}}=\frac{1}{2}C'{{V}^{2}}=\frac{1}{2}\left( \frac{C}{2} \right){{(2V)}^{2}}\] \[=C{{V}^{2}}\] Work done, \[W={{U}_{f}}-{{U}_{i}}\] \[=C{{V}^{2}}-\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}C{{V}^{2}}\]
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