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Engineering

Tell me? - Electrostatics - JEE Main-3

Let a total charge 2 Q be distributed in a sphere of radius R, with the charge density given by p(r)=kr, where r is the distance from the centre. Two charge A and B, of -Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then:

 

  • Option 1)

    a=8^{-1/4}R

     

     

     

     

  • Option 2)

    a=\frac{3R}{2^{1/4}}

  • Option 3)

    a=2^{-1/4}R

  • Option 4)

    a=R/\sqrt{3}

     

 
Answers (1)
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S solutionqc

by Gauss law

E\times4\pi a^{2}=\int _{o}^{a}\frac{Kr4\pi r^{2}}{\varepsilon _{0}}dr

E=\frac{Ka^{2}}{4\varepsilon _{0}}

force QE = \frac{kQa^{2}}{4\varepsilon _{0}}\cdots \cdots (1)

2Q=\int _{o}^{R}Kr\pi r^{2}dr\Rightarrow K=\frac{2Q}{\pi R^4}\cdots \cdots (2)

From (1) and (2)  QE = \frac{2Q}{\pi R^{4}}\times \frac{Qa^{2}}{4\varepsilon _{0}}\cdots \cdots (3)

also QE=\frac{1}{4\pi \varepsilon _{0}}\frac{QQ}{(2a)^{2}}\cdots \cdots (4)

from (3) and (4)         a=R 8^{-1/4}


Option 1)

a=8^{-1/4}R

 

 

 

 

Option 2)

a=\frac{3R}{2^{1/4}}

Option 3)

a=2^{-1/4}R

Option 4)

a=R/\sqrt{3}

 

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