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)

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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