A parallel plate capacitor without any dielectric within its plates, has a capacitance C, and is connected to a battery of emf V. The battery is disconnected and the plates of the capacitor are pulled apart until the separation between the plates is doubled. What is the work done by the agent pulling the plates apart, in this process?

- a)
- b)
- c)
- d)CV
^{2}

Correct answer is option 'A'. Can you explain this answer?

Stepway Academy
Feb 11, 2022 |

Capacitance of a parallel plate capacitor is C = ε_{0}A/d ……(i)

where A is the area of each plate and d is the distance between the plates. Initial energy stored in the capacitor,

When the separation between the plates is doubled , its capcaitance becomes

As the bettary is disconnected , so charged capacitor becomes isolated and charge on it will reamain constant ,

∴ Q = Q

C'V = C'V (AsQ = CV)

Final energy stored in the capacitor ,

Required work done,

where A is the area of each plate and d is the distance between the plates. Initial energy stored in the capacitor,

When the separation between the plates is doubled , its capcaitance becomes

As the bettary is disconnected , so charged capacitor becomes isolated and charge on it will reamain constant ,

∴ Q = Q

C'V = C'V (AsQ = CV)

Final energy stored in the capacitor ,

Required work done,

Capacitance of a parallel plate capacitor is C = ε0A/d ……(i)where A is the area of each plate and d is the distance between the plates. Initial energy stored in the capacitor,When the separation between the plates is doubled , its capcaitance becomesAs the bettary is disconnected , so charged capacitor becomes isolated and charge on it will reamain constant ,∴ Q = QCV =CV (AsQ = CV)Final energy stored in the capacitor ,Required work done,

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