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New Capacitance in Dielectric

Question

A capacitor has some dielectric between its plates, and the capacitor is connected to a dc source. The battery is now disconnected and then the dielectric is removed, then
capacitance will increase.
energy stored will decrease.
electric field will increase.
Voltage will decrease.

A

Voltage will decrease.

B

electric field will increase.

C

capacitance will increase.

D

energy stored will decrease.

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Solution

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When the capacitor is connected to dc source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric, capacitance decreases.
Energy stored $(u = \frac{q^{2}}{2 C})$ will increase
Potential $(V = \frac{q}{C})$ will also increase
Electric field $(E = \frac{V}{D})$ will increase

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