What is the escape speed of an electron launched from the surface of a 1.0-cm-diameter glass sphere that has been charged to 10 nC?

the minimum speed required for an particle to escape the external force can be defined as escape speed.

Here in this situation the energy required for the electron to escape the glass sphere is;

E=qV.

q is the charge of electron and V is the potential of the electron at glass sphere.

Now the radius of the Sphere is 0.005 m

The potential V can be denoted as;

$v=\frac{1}{4\pi {\in }_0}x\frac{q}{r}$

after substituting 9x10⁹ Nm²/C² is place of $\frac{1}{4\pi {\in }_0}$

this equation can be denoted as;

$$\begin{gathered} V=\frac{\left(9\times {10}^{9}N.m^2/C^2\right)\left(10nC\left({\displaystyle \frac{{10}^{-9}C}{1nC}}\right)\right)}{0.005} \\ =18000V \end{gathered}$$

Here 10 nC considered as q, and 0.005 m for r.

As per the above potential escape energy for the electron will be; $$\begin{gathered} E=qV \\ =(1.6\times {10}^{-19}C)\times 18000V \\ =2.88\times {10}^{-15}J \end{gathered}$$

As this amount of kinetic energy required by electron to escape the energy filed thus it can be considered as ;

$E=\frac{1}{2}m{v}^2$

Hence;

$$\begin{gathered} v=\sqrt{\frac{2E}{m}} \\ =\sqrt{\left(\frac{2\left(2.88\times {10}^{-15}J\right)}{9.1\times {10}^{-31}kg}\right)}\left[As,E=2.88\times {10}^{-15}J\right] \\ =7.9\times {10}^{7}m/s \end{gathered}$$

thus the escape speed required for electron is$7.9\times {10}^{7}m/s$