Electrostatics Question 150
Question: A parallel plate capacitor having a plate separation of 2 mm is charged by connecting it to a 300 V supply. The energy density is [BHU 2004]
Options:
A) $ 0.0\text{1}J/{{m}^{\text{3}}} $
B) $ 0.\text{1}J/{{m}^{\text{3}}} $
C) $ \text{1}.0J/{{m}^{\text{3}}} $
D) $ \text{1}0J/{{m}^{\text{3}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
The energy density of parallel plate capacitor is given by $ U=\frac{1}{2}{\varepsilon _{0}}E^{2}=\frac{1}{2}{\varepsilon _{0}}{{( \frac{V}{d} )}^{2}} $
$ =\frac{1}{2}\times 8.85\times {{10}^{-12}}C^{2}/Nm^{2}\times {{( \frac{300volt}{2\times {{10}^{-3}}m} )}^{2}} $
$ =0.1J/m^{3} $
