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 $
Show Answer
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} ] $
