Q11 of 805 Page 1

(a) How many electrons must be added to one plate and removed from the other so as to store 25.0 J of energy in a 5.0 nF parallel plate capacitor?

(b) How would you modify this capacitor so that it can store 50.0 J of energy without changing the charge on its plates?

(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 × 10-4coulomb


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 × 1015


Therefore number of electrons to be added to one plate and removed from an other plate are = 3.125 × 1015


(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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