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

Show Answer

Answer:

Correct Answer: 27. (a)

Solution:

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

As uncharged capacitor is connected parallel So, $C^{\prime}=2 C$

and $V_c=\frac{q_1+q_2}{C_1+C_2}$

$ \begin{align*} V_c & =\frac{q+0}{C+C}=\frac{CV}{C+C} \quad[\because q=cv] \\ \Rightarrow \quad V_c & =\frac{V}{2} \tag{i} \end{align*} $

Initial Energy of system, $U_i=\frac{1}{2} CV^{2}$

Final energy of system, $U_f=\frac{1}{2}(2 C)(\frac{V}{2})^{2}$

$=\frac{1}{2} CV^{2}(\frac{1}{2})$

From equation (i) and (ii)

$U_f=\frac{1}{2} U_i$

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