Question

# The electric field in a region is radially outward and is given by E=250 r V/m at a point (where r is the distance of the point from the origin). Calculate the charge contained in a sphere of radius 20 cm centred at the origin.

A
2.22×106 C
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B
2.22×108 C
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C
2.22×1010 C
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D
Zero
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Solution

## The correct option is C 2.22×10−10 C Here, E=250 r At the surface, r=R=0.02 m ⇒E=250×R=250×20×10−2 V/m Since electric field is radially outwards, the value of ∮→E.d→A at the Gaussian surface is : ∮→E.d→A=EA ⇒∮→E.d→A=E(4πR2) .....(i) On applying Gauss`s law, ϕnet=qinsideε0 ⇒∮→E⋅d→A=qinsideε0 (∵ϕnet=∮→E⋅d→A) Substituing from Eq.(i), ⇒E(4πR2)=qinsideε0 ⇒(250×20×10−2)×(20×10−2)2=qinside4πε0 ⇒50×400×10−4=(9×109) qinside ⇒qinside=29×109=209×10−10 ∴ qinside=2.22×10−10 C This is the enclosed charge in the given sphere. Why this question?Tip: Choosing a spherical Gaussian surface at a distanceof R=20 cm from origin will enclose all the containedcharge and radial direction of→Eat spherical surfacewill yield ∮→E⋅d→A=EA=E(4πR2).

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