Question

The electrical potential on the surface of a sphere of radius r due to a charge $3\times {{10}^{-6}}C$ is 500 V. The intensity of electric field on the surface of the sphere is $\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \right]$ $(in\text{ }N{{C}^{-1}})$ :

• A. $<10$
• B.

  20

• C. Between 10 and 20
• D. <5

Answer 1

Answer: (A)

$<10$

Answer 2

${{V}_{s}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{R}$ $\therefore$ $500=\frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}}{R}$ $\Rightarrow$ $R=\frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}}{500}$ $\Rightarrow$ $R=54\,\,m$ $\therefore$ Electric field on the surface ${{E}_{s}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{{{R}^{2}}}$ $=\frac{{{V}_{S}}}{R}=\frac{500}{54}=9.25<10\,N{{C}^{-1}}$