Electrostatics Ques 139
- A capacitor with capacitance $5 \mu \mathrm{F}$ is charged to $5 \mu \mathrm{C}$. If the plates are pulled apart to reduce the capacitance to $2 \mu \mathrm{F}$, how much work is done?
(a) $6.25 \times 10^{-6} \mathrm{~J}$
(b) $2.16 \times 10^{-6} \mathrm{~J}$
(c) $2.55 \times 10^{-6} \mathrm{~J}$
(d) $3.75 \times 10^{-6} \mathrm{~J}$
(Main 2019, 9 April I)
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Answer:
Correct Answer: 139.(d)
Solution:
Formula:
- Potential energy stored in a capacitor is
$ U=\frac{1}{2} Q V=\frac{1}{2} \frac{Q^{2}}{C} $
So, initial energy of the capacitor, $U_{i}=\frac{1}{2} Q^{2} / C_{1}$
Final energy of the capacitor, $U_{f}=\frac{1}{2} Q^{2} / C_{2}$
As we know, work done, $W=\Delta U=U_{f}-U_{i}$
$ =\frac{1}{2} Q^{2} \frac{1}{C_{2}}-\frac{1}{C_{1}} $
Here, $\quad Q=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}$,
$C_{1}=5 \mu \mathrm{F}=5 \times 10^{-6} \mathrm{~F}$,
$C_{2}=2 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F}$
$\Rightarrow \quad \Delta U=\frac{1}{2} \times\left(5 \times 10^{-6}\right)^{2} \frac{1}{2 \times 10^{-6}}-\frac{1}{5 \times 10^{-6}}$
$ \begin{aligned} & =\frac{1}{2} \times \frac{5 \times 5 \times 10^{-12}}{10^{-6}} \times \frac{3}{10} \\ & =\frac{25 \times 3}{20} \times 10^{-6} \mathrm{~J} \\ \Rightarrow \quad \Delta U & =3.75 \times 10^{-6} \mathrm{~J} \end{aligned} $
$\therefore$ Work done in reducing the capacitance from $5 \mu \mathrm{F}$ to $2 \mu \mathrm{F}$ by pulling plates of capacitor apart is $3.75 \times 10^{-6} \mathrm{~J}$.
BETA
