Answer :

(a) Given :

Energy to be stored in a capacitor (U) = 25 J

Capacitance (C) = 5 nF

We know that energy stored in a capacitor

where is Q = Charge in the capacitor

Q = 5 × 10^{-4}coulomb

We know that Q = ne

n = number of electrons

e = charge of an electron

5 × 10^{-4} = n × (1.6 × 10^{-19})

n = 3.125 × 10^{15}

Therefore number of electrons to be added to one plate and removed from an other plate are = 3.125 × 10^{15}

(b) Given:

U = 50 J

Q = 5 × 10^{-4} coulomb

From the formula

C = 2.5 × 10^{-9} F

The capacitance between two plates (C) = ϵA/d

where ϵ = absolute permittivity of the medium

A = Area of the plates

d = Distance between the two plates

C is reduced to half of it value when compared to the previous case.

As ϵ and Area of the plates are constant here, the distance between the plates should be doubled to get the desired value of capacitance.

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