Question: What is the escape speedfor an electron initially at rest on the surface of a sphere with a radius ofand a uniformly distributed charge of$\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{\text{C}}$? That is, what initial speed must the electron have in order to reach an infinite distance from the sphere and have zero kinetic energy when it gets there?

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

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, $U=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{R}$ (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:

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

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