An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal butopposite charge on its plates. All the geometric parameters of the capacitor (plate diameter andplate separation) are now DOUBLED. If the original energy stored in the capacitor was U0, howmuch energy does it now store?

U_f = (U_o)/2)

Explanation:

The capacitance of a given capacitor is given by the formula;

C = ε_o•A/d

While energy stored in plates capacitor is given as; U_o = Q²/2C

Now,we are told that that all the dimensions of the capacitor plate is doubled. Thus, we now have;

C' = ε_o•4A/2d

Hence, C' = 2C

so capacitance is now doubled

Thus, the final energy stored between the plates of capacitor is given as;

U_f = Q²/2C'

From earlier, we saw that C' = 2C.

Thus;

U_f = Q²/2(2C)

U_f = Q²/4C

Rearranging, we have;

U_f = (1/2)(Q²/2C)

From earlier, U_o = Q²/2C

Hence,

U_f = (1/2)(U_o)

Or

U_f = (U_o/2)

it will be pushed away from the source of the electric field. a negative charge in a cloud induces a positive charge in the ground below the cloud.

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