Question 19.3: Calculate the final speed of a free electron accelerated fro...

Conservation of energy states that

KE _{ i }+ PE _{ i }= KE _{ f }+ PE _{ f }.                 (19.17)

Entering the forms identified above, we obtain

q V=\frac{m v^{2}}{2}.                   (19.18)

We solve this for v :

v=\sqrt{\frac{2 q V}{m}}.                    (19.19)

Entering values for q, V, and m gives

v=\sqrt{\frac{2\left(-1.60 \times 10^{-19} C \right)(-100 J / C )}{9.11 \times 10^{-31} kg }}                 (19.20)

=5.93 \times 10^{6} m / s.

Discussion
Note that both the charge and the initial voltage are negative, as in Figure 19.4. From the discussions in Electric Charge and Electric Field, we know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. Those higher voltages produce electron speeds so great that relativistic effects must be taken into account. That is why a low voltage is considered (accurately) in this example.