The charge density of a spherical conductor of radius 50cm is 8.85×10−6C/m2. The electric field intensity in V/m at a point close to the surface will be :-
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Solution
Given,
Radius of sphere,r=0.5m
Charge density, qA=8.85×10−6Cm−2
From gauss law,
Electric field intensity, E=qAεo=8×10−68×10−12=106V/m