Two concentric coplanar rings are kept fixed in a gravity free region in y-z plane. One ring whose radius is R, carries a charge +Q and second ring whose radius is 4R, carries a charge of –8Q. Find the minimum speed with which a point charge of mass m and charge –q should be projected from a point at a distance 3R from the common center of rings on its axis so that it will reach to the center of the rings. Charges on the rings are distributed uniformly

Interesting point to note in this question is that at a point distance 3R charge –q will have repulsive force but on points close to the center it will have a attraction force. One should be careful while applying the energy conservation equation in these types of problems. First we have to find out a point where the electric field is zero because as beyond that, charge will have attraction force and will itself reach the center of the rings. Electric field at a point P distance x from the center.
 $E_p=\frac{1}{4\pi {\epsilon }_0}\frac{Qx}{{(R^2+x^2)}^{3/2}}-\frac{1}{4\pi {\epsilon }_0}\frac{8Qx}{{(16R^2+x^2)}^{3/2}}=0$
 $\Rightarrow x=2R$
So we have to give the charge enough KE so that it could reach this point because for x < 2R field will be attractive. Now we can apply work energy theorem  between these two points i.e. distance 3R and 2R.
 
Potential at A
 $=\frac{1}{4\pi {\epsilon }_0}[\frac{Q}{{(9R^2+R^2)}^{1/2}}-\frac{8Q}{{(9R^2+16R^2)}^{1/2}}]$
 $=-\frac{1}{4\pi {\epsilon }_0}\left(\frac{16-\sqrt{10}}{10}\right)\frac{Q}{R}$
Potential at point B
 $=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{{(4R^2+R^2)}^{1/2}}-\frac{1}{4\pi {\epsilon }_0}\frac{8Q}{{(4R^2+16R^2)}^{1/2}}=-\frac{1}{4\pi {\epsilon }_0}\frac{3}{\sqrt{5}}\frac{Q}{R}$
Now  $=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{{(4R^2+R^2)}^{1/2}}-\frac{1}{4\pi {\epsilon }_0}\frac{8Q}{{(4R^2+16R^2)}^{1/2}}=-\frac{1}{4\pi {\epsilon }_0}\frac{3}{\sqrt{5}}\frac{Q}{R}$
 $=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{R}\left[\frac{6\sqrt{5}+\sqrt{10}-16}{10}\right]$
 
$\Rightarrow v_{\min }={\left[\frac{2}{m}.\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{R}\frac{\{6\sqrt{5}+\sqrt{10}-16\}}{10}\right]}^{1/2}$