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Question

# A spherical volume contains a uniformly distributed charge of volume density 2.0×10−4 Cm−3. Find the strength of electric field at a point inside the volume at a distance 4.0 cm from the centre.

A

3×104 NC

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B

3×105 NC

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C

3×106 NC

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D

3×107 NC

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Solution

## The correct option is B 3×105 NC Consider a spherical surface of radius 4 cm as a Gaussian surface The electric field intensity at all points on the Gaussian surface should have the same magnitude and its direction radially outwards at each point as there cannot be any preferred direction other than radial. {Symmetry and uniform charge distribution} Let the magnitude of electric field strength be E. According to Gauss’ law, ∮→E.→ds=φ=qinϵ0 Again considering radially outward direction of elementary normal at each point on the Gaussian surface, we can claim, Angle between →E at →ds each small area element = 00 ⇒E(4πr2)=qinϵ0=ρ(43)πr3ϵ0 Where is ρ the given volume charge density. E=ρr3ϵ0=2×10−4×43×100×8.85×10−12NC E=3×105 NC

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