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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. State whether the capacitance, the energy stored in it, electric field, charge stored, and the voltage will increase, decrease, or remain constant.

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Solution

We know on removal of dielectric, capacitance will decrease by (K) C′=CK Energy stored in a capacitor U=Q22C As charge remains same Uα1C Energy will increase. Potential difference across the plate of capacitor V=QC Vα1C So, potential difference increases. V=Ed As d remains constant, E will also increase. Final Answer: C will decrease. Energy stored will increase. Electric field will increase. Charge stored will remain the same. V will increase.

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