A sphere of radius $1 \text{ cm}$ has potential of 8000Λ , …

Physics Electrostatics Potential & Capacitance Electric Field and Potential,Defference,Energy and Dipole Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: September 12, 2026

A sphere of radius $1 \text{ cm}$ has potential of 8000Λ , then energy density near its surface will be

$64 \times 10^{5} \mathrm{J} / \mathrm{m}^{3}$

$8 \times 10^{3} \, \mathrm{J/m^{3}}$

$32 J/m^{3}$

$2.83 \, \mathrm{J/m^3}$

Text Solution

Verified by Experts

The correct answer is:

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

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