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Question: A capacitor with capacitance \(5 \mu F\) is charged to \(5 \mu C\). If the plates are pulled apart t...

A capacitor with capacitance 5μF5 \mu F is charged to 5μC5 \mu C. If the plates are pulled apart to reduce the capacitance 2μF2\mu F, how much work is done:
A. 3.75×10−6J3.75 \times {10}^{-6}J
B. 2.55×10−6J2.55 \times {10}^{-6}J
C. 2.16×10−6J2.16 \times {10}^{-6}J
D. 6.25×10−6J6.25 \times {10}^{-6}J

Explanation

Solution

To solve this problem, find the difference between the initial and final potential energy of the capacitor. Use the formula for potential energy in terms of charge and capacitance. Substitute the values in the above-mentioned formula and find the initial and final potential energy of the capacitor. Subtract these potential energies to find how much work is done if the plates are pulled apart.
Formula used:
Work done= Uf−Ui{U}_{f}- {U}_{i}
U=Q22CU= \dfrac {{Q}^{2}}{2C}

Complete answer:
Given: C1=5μF{C}_{1}= 5 \mu F
C2=2μF{C}_{2}= 2 \mu F
Q=5μCQ= 5 \mu C
Work done= Uf−Ui{U}_{f}- {U}_{i} …(1)
Where, Ui{U}_{i} is the initial potential energy
Uf{U}_{f} is the final potential energy
Potential energy is given by,
U=Q22CU= \dfrac {{Q}^{2}}{2C}
Initial potential energy of the capacitor is given by,
Ui=Q22C1{U}_{i}= \dfrac {{Q}^{2}}{2{C}_{1}}
Substituting the values in above equation we get,
Ui=5×10−622×5×10−6{U}_{i}= \dfrac {{5 \times {10}^{-6}}^{2}}{2 \times 5 \times {10}^{-6}}
⇒Ui=2.5×10−6\Rightarrow {U}_{i}= 2.5 \times {10}^{-6} …(2)
Final potential energy of the capacitor is given by,
Uf=Q22C2{U}_{f}= \dfrac {{Q}^{2}}{2{C}_{2}}
Substituting the values in above equation we get,
Ui=5×10−622×2×10−6{U}_{i}= \dfrac {{5 \times {10}^{-6}}^{2}}{2 \times 2 \times {10}^{-6}}
⇒Ui=6.25×10−6\Rightarrow {U}_{i}= 6.25 \times {10}^{-6} …(3)
Substituting equation. (2) and (3) in equation. (1) we get,
W=6.25×10−6−2.5×10−6W= 6.25 \times {10}^{-6}-2.5 \times {10}^{-6}
⇒W=3.75×10−6\Rightarrow W= 3.75 \times {10}^{-6}
Hence, the work done is 3.75×10−6J3.75 \times {10}^{-6}J.

So, the correct answer is “Option A”.

Note:
Amount of charge a capacitor can store depends on two factors. First is the voltage and second is the physical characteristics of the capacitors. The capacitance depends on the dielectric constant of the medium between the plates, area of each plate and distance between the plates. When a dielectric medium is placed between the two parallel plates of the capacitor, it’s capacity of storing the energy increases. Its energy increases by a factor K which is called a dielectric constant. If you want to increase the capacitance of parallel plate capacitors then increase the area, decrease the separation between two plates and use a dielectric medium.