Aparallel-plate capacitor is connected to a battery. what happens to the stored energy if the plate separation is doubled while the capacitor remains connected to the battery?

(a) it remains the same
(b) it is doubles
(c) it decreases by a factor of 2
(d) it decreases by a factor of 4
(e) it increases by a factor of

(c)  As 'd' becomes doubled, energy decreases by the factor of 2

Explanation:

Energy stored in a parallel plate capacitor is given by:

As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.

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hope this

snell's law

refractve index = sin incidence / sin refraction