Electrostatics Question 390
Question: A sphere of radius $ 1cm $ has potential of $ 8000V $ , then energy density near its surface will be [RPET 1999]
Options:
A) $ 64\times 10^{5}J/m^{3} $
B) $ 8\times 10^{3}J/m^{3} $
C) $ 32J/m^{3} $
D) $ 2.83J/m^{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
Energy density $ u _{e}=\frac{1}{2}{\varepsilon _{0}}E^{2}=\frac{1}{2}\times 8.86\times {{10}^{-12}}\times {{( \frac{V}{r} )}^{2}} $
$ =2.83J/m^{3} $
