Question

. The plates of a charged parallel plate capacitor are brought closer after disconnecting the battery , the energy stored will
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Answer 1

Answer:

As battery is disconnected so charge q is constant.

The parallel plate capacitance is C= A∈o/d or C∝1/d

If capacitor are brought closer, d will decrease so c will increase.

Energy stored in capacitance is

U=q²/2c or U∝1/c

As C increases, so U will be decreases

The electric field between plates is E=q/A∈o

As q is constant so ∈ becomes unchanged.

potential difference between the plates s V=Ed or V∝d

As d decreases so V will decreases.

Answer 2

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