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} $



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