Charge Q = 5.00 $\mathit{\mu }\mathit{C}$C is distributed uniformly over the volume of an insulating sphere that has radius R = 12.0 cm. A small sphere with charge q = +3.00 $\mathit{\mu }\mathit{C}$C and mass$\mathbf{6}\mathbf{.}\mathbf{00}\mathit{x}{\mathbf{10}}^{\mathbf{5}}\mathbf{kg}$is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 cm of the surface of the large sphere?

$U_b=\frac{1}{4\mathrm{\pi }{€}_0}\frac{Qq}{r}$

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;

$U_b=\frac{1}{4\mathrm{\pi }{€}_0}\frac{Qq}{r}$

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;

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

The minimum speed (v) of the charge q;

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

Hence, the minimum speed is 150.0m/s.