Question

II. Compute the escape speed for an electron from the surface of a uniformly charged sphere of radius 1.22cm and total charge 1.76x10-15 C. Neglect gravitational forces.

Answer 1

The potential at the surface of the sphere is

$$V=\frac{Q}{4\pi\epsilon_0 R}.$$

The work required to remove the electron is

$$W=Vq=\frac{Qe}{4\pi\epsilon_0R}.$$

To escape from the sphere, the kinetic energy must be greater than this value (work-energy theorem):

$$E_k=\frac12mv^2=E_p,\\\space\\ \frac12mv^2=\frac{Qe}{4\pi\epsilon_0R},\\\space\\ v=\sqrt{\frac{Qe}{2\pi\epsilon_0mR}}=21.4\text{ km/s}.$$