Electrostatic Potential and Capacitance - Result Question 25

27. A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system : [2017]

(a) decreases by a factor of 2

(b) remains the same

(c) increases by a factor of 2

(d) increases by a factor of 4

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Answer:

Correct Answer: 27. (a)

Solution:

  1. (a) When battery is replaced by another uncharged capacitor

As uncharged capacitor is connected parallel So, C=2C

and Vc=q1+q2C1+C2

Vc=q+0C+C=CVC+C[q=cv](i)Vc=V2

Initial Energy of system, Ui=12CV2

Final energy of system, Uf=12(2C)(V2)2

=12CV2(12)

From equation (i) and (ii)

Uf=12Ui

i.e., Total electrostatic energy of resulting system decreases by a factor of 2



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