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)

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}

 

Preparation Products

Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
Buy Now
Test Series JEE Main Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions