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

Question

A sphere of radius 1 cm has potential of 8000 V, then energy density near its surface will be (A) 64 × 105 J / m 3 (B) 8 × 103 J / m 3 (C) 32 J / m 3 (D) 2.83 J / m 3

Tardigrade Admin · Accepted Answer

Energy density ue=(1/2) ε0 E2 =(1/2) × 8.86 × 10-12 ×((V/r))2 =2.83 J / m 3

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

$64 \times 1 0^{5} J / m^{3}$

$8 \times 1 0^{3} J / m^{3}$

$32 J / m^{3}$

$2.83 J / m^{3}$

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

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

64×105J/m3

8×103J/m3

32J/m3

2.83J/m3