Question

A sphere of radius $1cm$ has potential of $8000V$, then energy density near its surface will be____( in MKS )

• A. $31.87$
• B. $63.43$
• C. $8.52$
• D. $2.83$

Answer 1

Answer: (D)

$2.83$

Given,

$R=1cm=0.01m$

$V=8000V$

Electric field is given by

$E=\frac{V}{R}=\frac{8000}{0.01}=8\times {10}^{5}V/m$

The energy density near a surface of sphere,

$U=\frac{1}{2}{\epsilon }_0{E}^2$

$U=\frac{1}{2}\times 8.85\times {10}^{-12}\times 8\times 8\times {10}^{10}$

$U=2.83J/kgs$

The correct option is D.