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:
- (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
