# Homework Help: Velocity of an electron exam question

1. Oct 4, 2006

### eltel2910

An electron starts from rest 2.82 cm from the center of a uniformly charged sphere of radius 2.17 cm. If the sphere carries a total charge of 1.11×10-9 C, how fast will the electron be moving when it reaches the surface of the sphere?

This is an exam question that I got wrong and I am trying to figure out how to do it.

So it has something to do with figuring out the potential difference between the eletron's initial position and the surface of the sphere. But how do you do that? Then I know I can use the kinetic energy of the electron to find the velocity

2. Oct 4, 2006

### Andrew Mason

You have to express the field as a function of the electron's distance from the centre (inside the sphere) and then integrate from r = 0 to R (R being the radius of the sphere). As you have noted, the work done on the electron is the potential difference ($\int_0^R eEdr$) which is the electron's kinetic energy.

Use Gauss' law to determine the field, E as a function of r (the amount of enclosed charge is the charge density x volume of gaussian sphere of radius r).

AM

3. Oct 28, 2006

### lightarrow

More explicitly: The potential: V = q/4(pi)epsilon_0*r
substitute for the 2 values of r and find (delta)V
then you write (delta)V*q_e = kinetic energy = m*c^2/SQRT[1-(v/c)^2] - m*c^2
where q_e and m are the electron's charge and mass, and you find the speed v.

In this case you don't really have to use the relativistic formula, since, using the non-relativistic, it turns out that v/c is about 10^(-2).

Last edited: Oct 28, 2006
4. Oct 28, 2006

### Andrew Mason

If it is a uniformly charged sphere (charge density $\sigma = Q/V$), the enclosed charge at radius r (volume V(r)) is:

$$Q(r) = \sigma V(r) = \frac{4}{3}\pi r^3\sigma$$

Using Gauss' law and symmetry:

$$\oint E dA = 4\pi r^2E = Q(r)/\epsilon_0 = \frac{4}{3}\pi r^3\sigma /\epsilon_0$$

$$E = \frac{\sigma r}{3\epsilon_0}$$

The potential, therefore, is:

$$U = \int_0^R Edr = \frac{\sigma}{3\epsilon_0}\int_0^R rdr = \frac{\sigma}{6\epsilon_0}R^2$$

$$U = \frac{3Q}{4\pi R^3}\frac{R^2}{6\epsilon_0} = \frac{Q}{8R\pi\epsilon_0}$$

So the kinetic energy is:

$$KE_e = eU = \frac{1}{2}mv^2 = e\frac{Q}{8R\pi\epsilon_0}$$

$$v = \sqrt{e\frac{Q}{4mR\pi\epsilon_0}}$$

AM

Last edited: Oct 28, 2006