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**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: $\oint{\overrightarrow{E}}.d\overrightarrow{A}=\dfrac{{{Q}_{in}}}{{{\varepsilon }_{0}}}$

$F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

**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

${{F}_{AB}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}^{2}}}{{{\left( 2a \right)}^{2}}}.....\left( 2 \right)$

${{F}_{AB}}$ = electric force between A and B

k $=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$= 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,

$\oint{\overrightarrow{E}}.d\overrightarrow{A}=\dfrac{{{Q}_{in}}}{{{\varepsilon }_{0}}}$

$\overrightarrow{E}=$ Electric field

$dA$ = area

${{Q}_{in}}=$ Charge inside a sphere

${{\varepsilon }_{0}}$ = permittivity

Electric field is constant hence

$E\oint{dA=}{}_{0}\int{^{r}}\dfrac{\rho dv}{{{\varepsilon }_{0}}}$

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{align}

& \Rightarrow E.4\pi {{r}^{2}}={}_{0}\int{^{r}}kr\dfrac{4\pi {{r}^{2}}}{{{\varepsilon }_{0}}}dr \\

& \Rightarrow E.4\pi {{r}^{2}}=\dfrac{4\pi k}{{{\varepsilon }_{0}}}\left( \dfrac{{{r}^{4}}}{4} \right) \\

& \therefore E=\dfrac{k{{r}^{2}}}{4{{\varepsilon }_{0}}}....\left( 4 \right) \\

\end{align}$

Now we have to find k in order to do it we will use the below equation

Charge on whole sphere

$\begin{align}

& \Rightarrow {}_{o}\int{^{2Q}}dQ={}_{o}\int{^{R}}\rho dv \\

& \Rightarrow 2Q{{=}_{o}}\int{^{R}}kr4\pi {{r}^{2}}dr \\

& \Rightarrow 2Q=k4\pi \dfrac{{{R}^{4}}}{4} \\

& \therefore k=\dfrac{2Q}{\pi {{R}^{4}}} \\

\end{align}$

Now substitute the value of k in equation (4)

$E=\dfrac{Q{{r}^{2}}}{2\pi {{\varepsilon }_{0}}{{R}^{4}}}$

From figure we can put r = a

$E=\dfrac{Q{{a}^{2}}}{2\pi {{\varepsilon }_{0}}{{R}^{4}}}...\left( 5 \right)$

Now put all the values in equation (1)

$\begin{align}

& \Rightarrow {{F}_{AB}}={{F}_{AS}} \\

& \Rightarrow \dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}^{2}}}{{{\left( 2a \right)}^{2}}}=E.Q \\

\end{align}$

$\Rightarrow \dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}^{2}}}{{{\left( 4a \right)}^{2}}}=\dfrac{Q.{{a}^{2}}}{2\pi {{\varepsilon }_{0}}{{R}^{4}}}\times Q$

$\begin{align}

& \Rightarrow \dfrac{1}{16{{a}^{2}}}=\dfrac{{{a}^{2}}}{2{{R}^{4}}} \\

& \Rightarrow {{a}^{4}}=\dfrac{2}{16}{{R}^{4}} \\

& \Rightarrow {{a}^{4}}=\dfrac{1}{8}{{R}^{4}} \\

& \therefore a={{8}^{-\dfrac{1}{4}}}R \\

\end{align}$

**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.

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