Question

A sphere of radius $1$ cm has potential of $8000$V. The energy density near the surface of sphere will be?

• A. $64\times {10}^5$ $J/m^3$
• B. $8\times {10}^3$ $J/m^3$
• C. $32$ $J/m^3$
• D. $2.83$ $J/m^3$

Answer 1

The correct option is D $2.83$ $J/m^3$

Energy density = $=\left(\frac{1}{2}\right)\in 0E2$

$=\left(\frac{1}{2}\right)\times 8.86\times 10-12\times \left(\frac{v}{r}\right)2$

$=4.43\times 10-12\times \left[\right(8000\left)/\right(10-2\left)\right]2$

$=283.5\times 10-2$

$=2.83J/m3$