A) \[Q>{{Q}_{o}}\]
B) \[V>{{V}_{o}}\]
C) \[E>{{E}_{o}}\]
D) \[U>{{U}_{o}}\]
Correct Answer: A
Solution :
Capacitance will be increased when a dielectric is introduced in the capacitor but potential difference will remain the same because battery is still connected. So according to q = CV, charge will increase i.e. \[Q>{{Q}_{0}}\] and \[U=\frac{1}{2}Q{{V}_{0}},\,{{U}_{0}}=\frac{1}{2}{{Q}_{0}}{{V}_{0}}\] Þ \[Q>{{Q}_{0}}\] so \[U>{{U}_{0}}\]
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