Question

Let a total charge 2Q 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 charges A and B of –Q each are placed on diametrically opposite points, at equal distance a form the center. If A and B do not experience force, then:
$\begin{array}{rl}& A.a={\displaystyle \frac{3R}{2^{{\displaystyle \frac{1}{4}}}}}\\ & B.a={\displaystyle \frac{R}{\sqrt{3}}}\\ & C.a=8^{-{\displaystyle \frac{1}{4}}}R\\ & D.a=2^{-{\displaystyle \frac{1}{4}}}R\end{array}$

Answer 1

Hint: In order to find solution of this question force on A due to B must be equal to force on A due to sphere S to do so we have to find electric field on AB in order to do it we have to draw gauss’s surface inside a sphere S with radius r.
Formula used: $\oint \overrightarrow{E}.d\overrightarrow{A}={\displaystyle \frac{Q_{in}}{{\epsilon }_0}}$
$F={\displaystyle \frac{k{q}_1{q}_2}{r^2}}$

Complete answer:
In order to find a solution and equilibrium condition force on A due to B is equal to force on A due to S since A and B are at equal distance and have equal charge Q.
$F_{AB}=F_{AS}......\left(1\right)$
Now force on A due to B
$F_{AB}={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q^2}{{\left(2a\right)}^2}}.....\left(2\right)$
$F_{AB}$ = electric force between A and B
k $={\displaystyle \frac{1}{4\pi {\epsilon }_0}}$= proportionality constant
Q = charge
a = radius or distance between two charges.

Force due to S and A
$F_{AS}=E.Q.....\left(3\right)$
First we have to find electric field E we will take gauss’s surface to find electric field on A and B now,
$\oint \overrightarrow{E}.d\overrightarrow{A}={\displaystyle \frac{Q_{in}}{{\epsilon }_0}}$
$\overrightarrow{E}=$ Electric field
$dA$ = area
$Q_{in}=$ Charge inside a sphere
${\epsilon }_0$ = permittivity
Electric field is constant hence
$E\oint dA={}_0\int {}^r{\displaystyle \frac{\rho dv}{{\epsilon }_0}}$
Now we know that area of the radius r circle is
$A=4\pi r^2$
And it is given in question that $\rho =kr$
Now
$\begin{array}{rl}& \Rightarrow E.4\pi r^2={}_0\int {}^{r}kr{\displaystyle \frac{4\pi r^2}{{\epsilon }_0}}dr\\ & \Rightarrow E.4\pi r^2={\displaystyle \frac{4\pi k}{{\epsilon }_0}}\left({\displaystyle \frac{r^4}{4}}\right)\\ & \therefore E={\displaystyle \frac{k{r}^2}{4{\epsilon }_0}}....\left(4\right)\end{array}$
 Now we have to find k in order to do it we will use the below equation
Charge on whole sphere
$\begin{array}{rl}& \Rightarrow {}_o\int {}^{2Q}dQ={}_o\int {}^R\rho dv\\ & \Rightarrow 2Q{=}_o\int {}^{R}kr4\pi r^{2}dr\\ & \Rightarrow 2Q=k4\pi {\displaystyle \frac{R^4}{4}}\\ & \therefore k={\displaystyle \frac{2Q}{\pi R^4}}\end{array}$
Now substitute the value of k in equation (4)
$E={\displaystyle \frac{Q{r}^2}{2\pi {\epsilon }_0{R}^4}}$
From figure we can put r = a
$E={\displaystyle \frac{Q{a}^2}{2\pi {\epsilon }_0{R}^4}}...\left(5\right)$
Now put all the values in equation (1)
$\begin{array}{rl}& \Rightarrow F_{AB}=F_{AS}\\ & \Rightarrow {\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q^2}{{\left(2a\right)}^2}}=E.Q\end{array}$
$\Rightarrow {\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{Q^2}{{\left(4a\right)}^2}}={\displaystyle \frac{Q.a^2}{2\pi {\epsilon }_0{R}^4}}\times Q$
$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1}{16a^2}}={\displaystyle \frac{a^2}{2R^4}}\\ & \Rightarrow a^4={\displaystyle \frac{2}{16}}{R}^4\\ & \Rightarrow a^4={\displaystyle \frac{1}{8}}{R}^4\\ & \therefore a=8^{-{\displaystyle \frac{1}{4}}}R\end{array}$

Hence the correct option is (c).

Note:
In this question we have to consider both the charge A and B inside the sphere if we consider them outside the sphere the solution could lead us to the wrong answer or solution.