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

The Correct Option is A

$ {{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}} $