Trivial Problem With An Arrange

As charge density of sphere is $\rho_{(r)} = \dfrac{\rho_{0}}{r^2}$

So $\dfrac{dq}{dV} = \dfrac{\rho_{0}}{r^2}$

$dq =\dfrac{\rho_{0}}{r^2} \dfrac{dV}{dr}* dr$

$dq= \dfrac{\rho_{0}}{r^2} *4\pi r^2 dr$

$dq=4\pi \rho_{0}dr$

$Q = 4\pi \rho_{0} R$

So in equillibrium,

$\dfrac{kQ}{(2d-x)^2} = \dfrac{kq}{(x)^2}$

$\dfrac{\sqrt{Q}}{2d-x} = \dfrac{\sqrt{q}}{x}$

On solving,

$x(\sqrt{Q}+\sqrt{q}) = 2d\sqrt{q}$

$x=\dfrac{2d\sqrt{q}}{\sqrt{Q}+\sqrt{q}}$

$x=\dfrac{2d\sqrt{q}}{2\sqrt{\pi \rho_{0} R}+\sqrt{q}}$

So $a=b=2\Rightarrow a+b=4$ .

As the particle is at equillibrium the net force is $0$ . But the only forces acting are depending on the elcectric field which generates each fixed charge, which will have an opposite sign without depending on the sign of the charge we put between the fixed charges, so we expect that both of the electric fields are equal in magnitude:

By using Gauss' law: $\oint E_{q} ds =\frac{q}{\epsilon_{0}} \Rightarrow E_{q}=\frac{q}{4\pi x^2\epsilon_{0}}$ $\oint E_{s} ds = \frac{\displaystyle \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R}\frac{\rho_{0} r^2 sin(\phi) dr d\phi d\theta}{r^2}}{\epsilon_{0}}$ $E_{s}=\frac{4\pi R \rho_{0}}{4\pi (2d-x)^2 \epsilon_{0}}$

As they're equal in magnitude: $\frac{4\pi R \rho_{0}}{4\pi (2d-x)^2 \epsilon_{0}} = \frac{q}{4\pi x^2 \epsilon_{0}}$ $\frac{4\pi R \rho_{0}}{(2d-x)^2}=\frac{q}{x^2}$ $2\sqrt{\pi R \rho_{0}} x =\sqrt{q}(2d-x)$ $x=\frac{2d\sqrt{q}}{2\sqrt{\pi R \rho_{0}}+\sqrt{q}}$