91Ó°ÊÓ

$U_b=\frac{1}{4\mathrm{Ï€}{â‚\neg }_0}\frac{Qq}{r}$

The projected charge begins at point a, and as it moves towards a positive charge, the charge q has the greatest kinetic energy and zero potential energy at point a, where its speed will decrease due to the repulsive force between the two charges, so potential energy is zero.

The mechanical energy of the charge is conservative during this travel between points a and b, so the potential energy at point b is due to the electric field between the two charges and is given by;

$U_b=\frac{1}{4\mathrm{Ï€}{â‚\neg }_0}\frac{Qq}{r}$

R is the distance between the center and point b;

localid="1664262136493" $\begin{array}{l}r=R+8.00\mathrm{cm}\\ r=12.00\mathrm{cm}+8.00\mathrm{cm}=20.0\mathrm{cm}\end{array}$

Thus, the equation;

$\begin{array}{ll}{K}_a+U_a=K_b+u_b& \\ K_a-U_b& \\ \frac{1}{2}m{v}^2=\frac{1}{4\mathrm{Ï€\pm ð}}\frac{Qq}{r}& \end{array}$

The minimum speed (v) of the charge q;

localid="1664262189335" $$\begin{gathered} v=\sqrt{\frac{1}{4\mathrm{π}∈{}_0}\left[\frac{2\mathrm{Qq}}{\mathrm{mr}}\right]} \\ =\sqrt{\left(9.0×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left[\frac{2\left(5.00×{10}^{-6}\mathrm{C}×3.00×{10}^{-6}\mathrm{C}\right)}{\left(6.0×{10}^{-5}\mathrm{kg}\right)\left(0.20\mathrm{m}\right)}\right]} \end{gathered}$$

Hence, the minimum speed is 150.0m/s.