Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the +3.00 nC charge?

Mass of electron, m_e = 9.11 × 10^-31 kg.

Charge on electron, q_e = -1.60 ×10^-19 C.

Stationary charge 1, q₁ = 3.00 × 10^-9 C.

Stational charge 2, q₂ = 2.00 × 10^-9 C.

r_i = 0.250 m

r_1f = 0.100 m

r_2f = 0.400 m

The Law of energy conservation tells us that the total energy of a system is always conserved . It can not be gained or lost. It can change forms.

Initially, the charges are at the stationary position, so net initial energy is given by,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{i}}={\mathrm{U}}_{\mathrm{i}}+{\mathrm{K}}_{\mathrm{i}} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{{\mathrm{q}}_{\mathrm{e}}{\mathrm{q}}_1}{{\mathrm{r}}_{\mathrm{i}}}+\frac{{\mathrm{q}}_{\mathrm{e}}{\mathrm{q}}_2}{{\mathrm{r}}_{\mathrm{i}}}\right)+0 \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\left(-1.6\times {10}^{-19}\right)\left(3\times {10}^{-9}\right)}{0.250}+\frac{\left(-1.6\times {10}^{-19}\right)\left(2\times {10}^{-9}\right)}{0250}\right) \\ =-2.88\times {10}^{-17}\mathrm{J} \end{gathered}$$

Finally, the charge starts moving with a velocity, so net final energy is given by,

$$\begin{gathered} E_f=U_f+K_f \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q_e{q}_1}{r_{1f}}+\frac{q_e{q}_2}{r_{2f}}\right)+\frac{1}{2}{m}_e{v}_f^2 \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\left(-1.6\times {10}^{-19}\right)\left(3\times {10}^{-9}\right)}{0.100}+\frac{\left(-1.6\times {10}^{-19}\right)\left(2\times {10}^{-9}\right)}{0.400}\right)+\frac{1}{2}{m}_e{v}_f^2 \\ =-5.04\times {10}^{-17}J+\frac{1}{2}{m}_e{v}_f^2 \end{gathered}$$

Here, v_f is the velocity of the electron.

By Law of conservation of energy,

$$\begin{gathered} E_i=E_f \\ {U}_i+K_i=U_f+k_f \\ -2.80\times {10}^{-17}=-5.04\times {10}^{-17}+\frac{1}{2}{m}_e{v}_f^2 \\ {v}_f=\sqrt{\frac{2\left(-2.88\times {10}^{-17}+5.04\times {10}^{-17}\right)}{9.11\times {10}^{-31}}} \\ {v}_f=6.89\times {10}^{6}m/s \end{gathered}$$

Therefore, the speed of an electron when it is 10 cm away from q₁ is 6.8910⁶ m/s.