SATHEE: Electrostatics 5 Question 5

Electrostatics 5 Question 5

6. A capacitor with capacitance $5 μ F$ is charged to $5 μ C$ . If the plates are pulled apart to reduce the capacitance to $2 μ F$ , how much work is done?

(a) $6.25 \times 10^{- 6} J$

(b) $2.16 \times 10^{- 6} J$

(d) $3.75 \times 10^{- 6} J$

(Main 2019, 9 April I)

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Answer:

Correct Answer: 6. (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, $W = Δ U = U_{f} - U_{i}$

$= \frac{1}{2} Q^{2} \frac{1}{C_{2}} - \frac{1}{C_{1}}$

Here, $Q = 5 μ C = 5 \times 10^{- 6} C$ ,

$C_{1} = 5 μ F = 5 \times 10^{- 6} F$ ,

$C_{2} = 2 μ F = 2 \times 10^{- 6} F$

$\Rightarrow Δ U = \frac{1}{2} \times {(5 \times 10^{- 6})}^{2} \frac{1}{2 \times 10^{- 6}} - \frac{1}{5 \times 10^{- 6}}$

$\begin{aligned} = \frac{1}{2} \times \frac{5 \times 5 \times 10^{- 12}}{10^{- 6}} \times \frac{3}{10} \\ = \frac{25 \times 3}{20} \times 10^{- 6} J \\ \Rightarrow Δ U & = 3.75 \times 10^{- 6} J \end{aligned}$

$∴$ Work done in reducing the capacitance from $5 μ F$ to $2 μ F$ by pulling plates of capacitor apart is $3.75 \times 10^{- 6} J$ .

Electrostatics 5 Question 5

6. A capacitor with capacitance $5 μ F$ is charged to $5 μ C$ . If the plates are pulled apart to reduce the capacitance to $2 μ F$ , how much work is done?

Answer:

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Electrostatics 5 Question 5

6. A capacitor with capacitance 5μF is charged to 5μC. If the plates are pulled apart to reduce the capacitance to 2μF, how much work is done?

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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6. A capacitor with capacitance $5 μ F$ is charged to $5 μ C$ . If the plates are pulled apart to reduce the capacitance to $2 μ F$ , how much work is done?