The volume charge density of a sphere of radius $6 \mathrm{~m}$ is $2 \,\mu \mathrm{C} \,\mathrm{cm}^{-3}$. The number of lines of force per unit surface area coming out from the surface of the sphere is _______________ $\times 10^{10} \,\mathrm{NC}^{-1}$.
[Given : Permittivity of vacuum $$\epsilon_{0}=8.85 \times 10^{-12} \,\mathrm{C}^{2}\, \mathrm{~N}^{-1}-\mathrm{m}^{-2}$$ )
Answer (integer)
45
Solution
<p>$\rho$ = 2 $\mu$c/cm<sup>3</sup></p>
<p>R = 6 m</p>
<p>Number of lines of force per unit area = Electric field at surface.</p>
<p>$= {{KQ} \over {{R^2}}}$</p>
<p>$= {1 \over {4\pi {\varepsilon _0}}}{{\rho {4 \over 3}\pi {R^3}} \over {{R^2}}}$</p>
<p>$= {{\rho R} \over {3{ \in _0}}}$</p>
<p>$$ = {{2 \times {{10}^{ - 6}} \times {{10}^6} \times 6} \over {3 \times 8.85 \times {{10}^{ - 12}}}}$$</p>
<p>$= 0.45197 \times {10^{12}}$</p>
<p>$= 45.19 \times {10^{10}}$ N/C</p>
<p>$\simeq 45 \times {10^{10}}$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Field and Field Lines
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