Electrostatic Potential And Capacitance Ques 20
20. 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: 20.(a)
Solution:
- (a) When battery is replaced by another uncharged capacitor
As uncharged capacitor is connected parallel So, $C^{\prime}=2 C$
and $V_c=\frac{q_1+q_2}{C_1+C_2}$
$V_c =\frac{q+0}{C+C}=\frac{CV}{C+C} \quad[\because q=cv]$
$\Rightarrow \quad V_c =\frac{V}{2} $
Initial Energy of system, $U_i=\frac{1}{2} CV^{2}$ $\quad$ …….(i)
Final energy of system, $U_f=\frac{1}{2}(2 C)(\frac{V}{2})^{2}$
$=\frac{1}{2} CV^{2}(\frac{1}{2})$ $\quad$ …….(ii)
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 $2$
BETA
