Question

The distance between the plates of a parallel -plate capactior of capacitance C is double .Its new capacitance will be

Answer 1

With double seperation of the capacitor( condenser) plates it’s capacitance reduces to the half value as C is inversely proportional to the distance between plates. After charging there it contains energy. There are two situations arise when you separate plates after charging.

1). If the separation done after removing the battery, previous charges of the plates remain constant. W=Q.Q/2C, C changes to C/2 , then W changes to 2W, the additional W is the work to be done by physically. Plates attract one another, to separate them addiwork to be done.

2). If the separation done without removing the battery, changes occur by making ultimate plates voltage remain constant. So we have to explain it by using W= C.V.V/2, as C becomes C/2, W to be reduced. Hence there is certain amount of condenser discharge. So some work to be done physically to separate the plates. Simultaneously capacitor reduces energy by deschaging. We have to calculate and compare with figures which one is grater. Then only we can come to the conclusion which work to be done or energy release by the capacitor.

Answer 2

Explanation:

The capacitance of the parallel plate capacitor is given by :

$C=\dfrac{\epsilon_o A}{d}$

A is area of cross section of parallel plate capacitor

d is separation between plates

$\epsilon_o$ is permittivity of free space

It is clear that the capacitance is inversely proportional to the separation between plates. If the distance between the plates of a parallel-plate capacitor of capacitance C is double, its new capacitance gets halved.

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Parallel plate capacitor

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