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Correct option is D)

$E=ΔV/Δr$ ,

where $Δr$ is the distance between plates , which is constant .

Now potential energy of a capacitor is given by ,

$U=(1/2)qΔV$ ,

charge is given constant , therefore due to decrease in potential difference $ΔV$ , potential energy $U$ also decreases .

Solve any question of Electrostatic Potential and Capacitance with:-

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