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

 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 ${\text{KE}}_{i}=0,\phantom{\rule{0.25em}{0ex}}{\text{KE}}_{f}=½{\mathrm{mv}}^{2},\phantom{\rule{0.25em}{0ex}}{\text{PE}}_{i}=\mathrm{qV}{\text{, and PE}}_{f}=0.$

Solution

Conservation of energy states that

${\text{KE}}_{i}+{\text{PE}}_{i}{\text{= KE}}_{f}+{\text{PE}}_{f}\text{.}$

Entering the forms identified above, we obtain

$\text{qV}=\frac{{\text{mv}}^{2}}{\text{2}}\text{.}$

We solve this for $v$ :

$v=\sqrt{\frac{2\text{qV}}{m}}\text{.}$

Entering values for $q,\phantom{\rule{0.25em}{0ex}}V\text{, and}\phantom{\rule{0.25em}{0ex}}m$ gives

$\begin{array}{lll}v& =& \sqrt{\frac{2\left(–1.60×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{C}\right)\left(\text{–100 J/C}\right)}{9.11×{\text{10}}^{\text{–31}}\phantom{\rule{0.25em}{0ex}}\text{kg}}}\\ & =& 5.93×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m/s.}\end{array}$

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.

## Section summary

• Electric potential is potential energy per unit charge.
• The potential difference between points A and B, ${V}_{B}–{V}_{A}$ , defined to be the change in potential energy of a charge $q$ moved from A to B, is equal to the change in potential energy divided by the charge, Potential difference is commonly called voltage, represented by the symbol $\text{Δ}V$ .
$\Delta V=\frac{\text{ΔPE}}{q}\phantom{\rule{0.25em}{0ex}}\text{and ΔPE =}\phantom{\rule{0.25em}{0ex}}q\Delta V\text{.}$
• An electron volt is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,
$\begin{array}{lll}\text{1 eV}& =& \left(1.60×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{C}\right)\left(1 V\right)=\left(1.60×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{C}\right)\left(1 J/C\right)\\ & =& 1.60×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J.}\end{array}$
• Mechanical energy is the sum of the kinetic energy and potential energy of a system, that is, $\text{KE}+\text{PE}.$ This sum is a constant.

## Conceptual questions

Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?

If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be done without exerting a force? Explain.

What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy?

Voltages are always measured between two points. Why?

How are units of volts and electron volts related? How do they differ?

## Problems&Exercises

Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be $1\text{.}\text{67}×{\text{10}}^{–\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\text{.}$

42.8

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