91Ó°ÊÓ

Distance between two electrons,$d\text{'}=2\text{cm}$

Using the given concept and the values of kinetic and potential energies derived in both the initial and final case, we can get the required value of the speed of the electron by substituting these values in the equation of the conservation of energy.

Formulae:

The kinetic energy of a particle in motion, $KE=\frac{1}{2}m{v}^2$ (i)

The potential energy of a particle at a distance, $U=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{r}$ (ii)

According to energy conservation of a system, $U_i+K{E}_i=U_f+K{E}_f$ (iii)

As the electron is shot from infinity, the initial potential energy of the electron is given as: $U_i=0$

The equation for final electric potential energy on the shot electron due to other two electrons is given using equation (ii) as follows:

$$\begin{gathered} U_f=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{d}+\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{d} \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q^2}{d} \end{gathered}$$

Now, the initial kinetic energy of the third electron is given using equation (i) as:

role="math" localid="1662098067495" $K{E}_i=\frac{1}{2}m{v}^2$

Here, m is the mass of the electron and is the speed of the electron.

Again, the final kinetic energy of the electron is zero as it momentarily stops st the midway.

$U_i+K{E}_i=U_f+K{E}_f$Apply conservation of energy from the initial and final states of three electron systems.

The distance between the mid electron and any other side of the electron is given as:

$$\begin{gathered} d=\frac{2.0\text{cm}}{2} \\ =1.0\text{cm} \\ =1.0×{10}^{-2}\text{m} \end{gathered}$$

Thus, using these given values in equation (iii), we can get the initial speed of the electron as follows:

$$\begin{gathered} 0+K{E}_i=U_f+0 \\ \frac{1}{2}m{v}^2=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{2q^2}{d} \\ v=\sqrt{\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\right)\frac{4q^2}{md}} \\ =\sqrt{\left(8.99×{10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\frac{4{(1.6×{10}^{-19}\text{C})}^2}{\left(9.11×{10}^{-31}\text{kg}\right)\left(1.0×{10}^{-2}\text{m}\right)}\right)} \\ =0.320\text{km/s} \end{gathered}$$

Hence, the value of the initial speed is 0.320 km/s.