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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 2.22×1010 C

Here, E=250 r

At the surface, r=R=0.02 m

E=250×R=250×20×102 V/m

Since electric field is radially outwards, the value of E.dA at the Gaussian surface is :

E.dA=EA

E.dA=E(4πR2) .....(i)

On applying Gauss`s law,

ϕnet=qinsideε0

EdA=qinsideε0

(ϕnet=EdA)

Substituing from Eq.(i),

E(4πR2)=qinsideε0

(250×20×102)×(20×102)2=qinside4πε0

50×400×104=(9×109) qinside

qinside=29×109=209×1010

qinside=2.22×1010 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 ofEat spherical surfacewill yield EdA=EA=E(4πR2).

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