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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorForce & Energy analysisMedium2 minPYQ_2008

PhysicsMediumsingle choice

The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section $A$ such that the uniform electric field between the plates is $E$ , isNEET 2008 (Screening)

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Answer:

Solution:

Energy given by the cell $E = C V^{2}$ Here, $C =$ capacitance of condenser $= \frac{A ε _{0}}{d}$ $V =$ potential difference across the plates $= E d$ Therefore, $E = (\frac{A ε _{0}}{d}) (E d)^{2} = A ε_{0} E^{2} d$

Stream:NEETSubject:PhysicsTopic:CapacitorSubtopic:Force & Energy analysis

⏱ 2mℹ️ Source: PYQ_2008

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