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  4. A sphere of radius 1 cm has potential of 8000 V, then energy density near its surface will be

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Q. A sphere of radius $1\, cm$ has potential of $8000\, V$, then energy density near its surface will be

Electrostatic Potential and Capacitance

Solution:

Energy density $u_{e}=\frac{1}{2} \varepsilon_{0} E^{2}$
$=\frac{1}{2} \times 8.86 \times 10^{-12} \times\left(\frac{V}{r}\right)^{2}$
$=2.83\, J / m ^{3}$