Electrostatics Question 809

Question: The energy required to change a parallel plate condenser of plate separation d and plated area of cross-section A such that the uniform electric field between the plates is E, is

Options:

A) $ {\in _{0}}E^{2}Ad $

B) $ \frac{1}{2}{\in _{0}}E^{2}Ad $

C) $ \frac{1}{2}{\in _{0}}E^{2}/Ad $

D) $ {\varepsilon _{0}}E^{2}/Ad $

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Answer:

Correct Answer: A

Solution:

[a] Energy required to charge the capacitor is $ W=U=QV $

$ \Rightarrow U=CV^{2}=\frac{{\varepsilon _{0}}A}{d}y^{2}=\frac{{\varepsilon _{0}}Ad}{d^{2}} $

$ V^{2}={\varepsilon _{0}}E^{2}Ad\text{ }[ \because E=\frac{V}{d} ] $



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