Electrostatics 5 Question 5
6. A capacitor with capacitance $5 \mu F$ is charged to $5 \mu C$. If the plates are pulled apart to reduce the capacitance to $2 \mu F$, how much work is done?
(a) $6.25 \times 10^{-6} J$
(b) $2.16 \times 10^{-6} J$
(c) $2.55 \times 10^{-6} J$
(d) $3.75 \times 10^{-6} J$
(Main 2019, 9 April I)
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Answer:
Correct Answer: 6. (b, d)
Solution:
- 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, $\quad Q=5 \mu C=5 \times 10^{-6} C$,
$C _1=5 \mu F=5 \times 10^{-6} F$,
$C _2=2 \mu F=2 \times 10^{-6} F$
$\Rightarrow \quad \Delta U=\frac{1}{2} \times\left(5 \times 10^{-6}\right)^{2} \frac{1}{2 \times 10^{-6}}-\frac{1}{5 \times 10^{-6}}$
$$ \begin{aligned} & =\frac{1}{2} \times \frac{5 \times 5 \times 10^{-12}}{10^{-6}} \times \frac{3}{10} \\ & =\frac{25 \times 3}{20} \times 10^{-6} J \\ \Rightarrow \quad \Delta U & =3.75 \times 10^{-6} J \end{aligned} $$
$\therefore$ Work done in reducing the capacitance from $5 \mu F$ to $2 \mu F$ by pulling plates of capacitor apart is $3.75 \times 10^{-6} J$.
