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
  • a)
    capacitance will increase.
  • b)
    energy stored will decrease.
  • c)
    electric field will increase.
  • d)
    voltage will decrease.
Correct answer is option 'C'. Can you explain this answer?

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Answers

Vp Classes
Feb 09, 2022
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 decrease.
Energy stored (u = q2/2C) will increase.
Potential (V = q/C) will also increase.
Electric field (E = V/D) will increase.

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 decrease.Energy stored (u = q2/2C) will increase.Potential (V = q/C) will also increase.Electric field (E = V/D) will increase.
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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 decrease.Energy stored (u = q2/2C) will increase.Potential (V = q/C) will also increase.Electric field (E = V/D) will increase.