Electric potential energy: potential difference  (Page 5/10)

Electrical potential energy converted to kinetic energy

Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to three significant figures.)

Strategy

We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be ${\text{KE}}_{\mathrm{i}}=\mathrm{0,}{\text{KE}}_{\mathrm{f}}=\mathrm{½}{\mathrm{mv}}^2,{\text{PE}}_{\mathrm{i}}=\mathrm{qV}{\text{, and PE}}_{\mathrm{f}}=0.$

Solution

Conservation of energy states that

Entering the forms identified above, we obtain

We solve this for $v$ :

Entering values for $q,V\text{, and}m$ gives

Discussion

Note that both the charge and the initial voltage are negative, as in [link] . 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.