JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 25)
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 $$\epsilon_{0}=8.85 \times 10^{-12} \,\mathrm{C}^{2}\, \mathrm{~N}^{-1}-\mathrm{m}^{-2}$$ )
Explanation
$$\rho$$ = 2 $$\mu$$c/cm3
R = 6 m
Number of lines of force per unit area = Electric field at surface.
$$ = {{KQ} \over {{R^2}}}$$
$$ = {1 \over {4\pi {\varepsilon _0}}}{{\rho {4 \over 3}\pi {R^3}} \over {{R^2}}}$$
$$ = {{\rho R} \over {3{ \in _0}}}$$
$$ = {{2 \times {{10}^{ - 6}} \times {{10}^6} \times 6} \over {3 \times 8.85 \times {{10}^{ - 12}}}}$$
$$ = 0.45197 \times {10^{12}}$$
$$ = 45.19 \times {10^{10}}$$ N/C
$$ \simeq 45 \times {10^{10}}$$
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