91Ó°ÊÓ

1. The electron is initially at rest.
2. Radius of the sphere,$R=0.01\text{m}$
3. Charge on the sphere,$q=1.6×{10}^{-15}\text{C}$
4. Final kinetic energy,$K.E_f=0$
5. The charge of the electron,$q_2=1.6×{10}^{-19}\text{C}$

Using the concept of the energy of a conducting shell, we can get the required speed of the electron by using the law of conservation of energy. Here, we get an expression; by solving it we get the answer of the speed from the given values.

Formulae:

The potential energy of a conducting shell, $U=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{R}$ (i)

The kinetic energy of a body in motion, $K=\frac{1}{2}m{v}^2$ (ii)

According to the law of conservation of energy, $(K.E.{)}_i+(P.E.{)}_i=(K.E.{)}_f+(P.E.{)}_f$ (iii)

Using the given data and equations (i) and (ii) in the equation (iii), we can get the escape speed of the electron as follows:

$$\begin{gathered} K_f-K_i=-U \\ 0-\frac{1}{2}m{v_i}^2=-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{R} \\ \frac{1}{2}m{v_i}^2=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q_1{q}_2}{R} \\ {v}_i=\sqrt{\frac{2×9.0×{10}^9×1.6×{10}^{-15}×1.6×{10}^{-19}}{9.1×{10}^{-31}×0.01}} \\ =2.25×{10}^4\text{m/s} \\ =22\text{km/s} \end{gathered}$$

Hence, the escape speed of the electron is 22 km/s.