Electrostatics

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

1) $64 \times {10^5}J/{m^3}$

2) $8 \times {10^3}J/{m^3}$

3) $32\,J/{m^3}$

4) $2.83\,J/{m^3}$

Solution:

Explanation: Explanation

electric-field-potential

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)