The Electron-Volt

The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. The particle may do its damage by direct collision, or it may create harmful X-rays, which can also inflict damage. It is useful to have an energy unit related to submicroscopic effects.

Figure 7.13 shows a situation related to the definition of such an energy unit. An electron is accelerated between two charged metal plates, as it might be in an old-model television tube or oscilloscope. The electron gains kinetic energy that is later converted into another form—light in the television tube, for example. (Note that in terms of energy, “downhill” for the electron is “uphill” for a positive charge.) Since energy is related to voltage by $\text{Δ}U=q\text{Δ}V$, we can think of the joule as a coulomb-volt.

Figure 7.13 A typical electron gun accelerates electrons using a potential difference between two separated metal plates. By conservation of energy, the kinetic energy has to equal the change in potential energy, so $KE=qV$. The energy of the electron in electron-volts is numerically the same as the voltage between the plates. For example, a 5000-V potential difference produces 5000-eV electrons. The conceptual construct, namely two parallel plates with a hole in one, is shown in (a), while a real electron gun is shown in (b).

An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electron accelerated through 50 V gains 50 eV. A potential difference of 100,000 V (100 kV) gives an electron an energy of 100,000 eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V gains 200 eV of energy. These simple relationships between accelerating voltage and particle charges make the electron-volt a simple and convenient energy unit in such circumstances.

The electron-volt is commonly employed in submicroscopic processes—chemical valence energies and molecular and nuclear binding energies are among the quantities often expressed in electron-volts. For example, about 5 eV of energy is required to break up certain organic molecules. If a proton is accelerated from rest through a potential difference of 30 kV, it acquires an energy of 30 keV (30,000 eV) and can break up as many as 6000 of these molecules $(\mathrm{30,000}\text{eV}÷5\text{eV}\text{per molecule}=6000\text{molecules).}$ Nuclear decay energies are on the order of 1 MeV (1,000,000 eV) per event and can thus produce significant biological damage.

Conservation of Energy

The total energy of a system is conserved if there is no net addition (or subtraction) due to work or heat transfer. For conservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.

Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, $K+U=\text{constant}\text{.}$ A loss of U for a charged particle becomes an increase in its K. Conservation of energy is stated in equation form as

or

where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem solving.

Voltage and Electric Field

So far, we have explored the relationship between voltage and energy. Now we want to explore the relationship between voltage and electric field. We will start with the general case for a non-uniform $\overrightarrow{\text{E}}$ field. Recall that our general formula for the potential energy of a test charge q at point P relative to reference point R is

When we substitute in the definition of electric field $(\overrightarrow{\text{E}}=\overrightarrow{\text{F}}\text{/}q),$ this becomes

Applying our definition of potential $(V=U\text{/}q)$ to this potential energy, we find that, in general,

From our previous discussion of the potential energy of a charge in an electric field, the result is independent of the path chosen, and hence we can pick the integral path that is most convenient.

Consider the special case of a positive point charge q at the origin. To calculate the potential caused by q at a distance r from the origin relative to a reference of 0 at infinity (recall that we did the same for potential energy), let $P=r$ and $R=\infty ,$ with $d\overrightarrow{\text{l}}=d\overrightarrow{\text{r}}=\widehat{\text{r}}dr$ and use $\overrightarrow{\text{E}}=\frac{kq}{r^2}\widehat{\text{r}}.$ When we evaluate the integral

for this system, we have

which simplifies to

This result,

is the standard form of the potential of a point charge. This will be explored further in the next section.

To examine another interesting special case, suppose a uniform electric field $\overrightarrow{\text{E}}$ is produced by placing a potential difference (or voltage) $\text{Δ}V$ across two parallel metal plates, labeled A and B (Figure 7.14). Examining this situation will tell us what voltage is needed to produce a certain electric field strength. It will also reveal a more fundamental relationship between electric potential and electric field.

Figure 7.14 The relationship between V and E for parallel conducting plates is $E=\frac{-\text{Δ}V}{d}$. The field points toward lower potential V.

From a physicist’s point of view, either $\text{Δ}V$ or $\overrightarrow{\text{E}}$ can be used to describe any interaction between charges. However, $\text{Δ}V$ is a scalar quantity and has no direction, whereas $\overrightarrow{\text{E}}$ is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field, a scalar quantity, is represented by E.) The relationship between $\text{Δ}V$ and $\overrightarrow{\text{E}}$ is revealed by calculating the work done by the electric force in moving a charge from point A to point B. But, as noted earlier, arbitrary charge distributions require calculus. We therefore look at a uniform electric field as an interesting special case.

The work done by the electric field in Figure 7.14 to move a positive charge q from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

Work is $W=\overrightarrow{\text{F}}·\overrightarrow{\text{d}}=Fd\text{cos}\theta$; here $\text{cos}\theta =1$, since the path is parallel to the field. Thus, $W=Fd$. Since $F=qE$, we see that $W=qEd$.

Substituting this expression for work into the previous equation gives

The charge cancels, so we obtain for the voltage between points A and B

where d is the distance from A to B, or the distance between the plates in Figure 7.14. Note that this equation implies that the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus, the following relation among units is valid:

Furthermore, we may extend this to the integral form. Substituting Equation 7.6 into our definition for the potential difference between points A and B, we obtain

which simplifies to

The potential difference is negative (V is lower at B than at A) when the displacement is in the same direction as the field. In other words, the electric field points toward lower electric potential. We are often only interested in the magnitude of the electric field, in which case you may see $\text{Δ}V=Ed$ instead of $\left|\text{Δ}V\right|=\left|Ed\right|$ and it is understood that all of the variables in the equation represent the magnitudes of the quantities. As a demonstration, from this we may calculate the potential difference between two points (A and B) equidistant from a point charge q at the origin, as shown in Figure 7.15.

Figure 7.15 The arc for calculating the potential difference between two points that are equidistant from a point charge at the origin.

To do this, we integrate around an arc of the circle of constant radius r between A and B, which means we let $d\overrightarrow{\text{l}}=r\widehat{\phi }d\phi ,$ while using $\overrightarrow{\text{E}}=\frac{kq}{r^2}\widehat{\text{r}}$. Thus,

for this system becomes

However, $\widehat{\text{r}}·\widehat{\phi }=0$ and therefore

This result, that there is no difference in potential along a constant radius from a point charge, will come in handy when we map potentials.

Example 7.7

What Is the Highest Voltage Possible between Two Plates?

Dry air can support a maximum electric field strength of about $3.0\times {10}^6\text{V/m}\text{.}$ Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?

Strategy

We are given the maximum electric field E between the plates and the distance d between them. We can use the equation $\text{Δ}V=Ed$ (recall that these represent absolute values) to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

$$\text{Δ}V=Ed.$$

Entering the given values for E and d gives

$$\text{Δ}V=(3.0\times {10}^6\text{V/m})(0.025\text{m})=7.5\times {10}^4\text{V}$$

or

$$\text{Δ}V=75\text{kV}\text{.}$$

(The answer is quoted to only two digits, since the maximum field strength is approximate.)

Significance

One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5-cm (1-in.) gap, or 150 kV for a 5-cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage can cause a spark if there are spines on the surface, since sharp points have larger field strengths than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up with static electricity on dry days (Figure 7.16).

Figure 7.16 A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between adjacent plates is not high enough to cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). This form of detector is now archaic and no longer in use except for demonstration purposes. (credit b: modification of work by Jack Collins)

Example 7.9

Calculating Potential of a Point Charge

Given a point charge $q=+2.0\text{nC}$ at the origin, calculate the potential difference between point $P_1$ a distance $a=4.0\text{cm}$ from q, and $P_2$ a distance $b=12.0\text{cm}$ from q, where the two points have an angle of $\phi =24\text{°}$ between them (Figure 7.17).

Figure 7.17 Find the difference in potential between $P_1$ and $P_2$.

Strategy

Do this in two steps. The first step is to use $V_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ and let $A=a=4.0\text{cm}$ and $B=b=12.0\text{cm},$ with $d\overrightarrow{\text{l}}=d\overrightarrow{\text{r}}=\widehat{\text{r}}dr$ and $\overrightarrow{\text{E}}=\frac{k_{e}q}{r^2}\widehat{\text{r}}.$ Then perform the integral. The second step is to integrate $V_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ around an arc of constant radius r, which means we let $d\overrightarrow{\text{l}}=r\widehat{\phi }d\phi$ with limits $0\le \phi \le 24\text{°},$ still using $\overrightarrow{\text{E}}=\frac{k_{e}q}{r^2}\widehat{\text{r}}$. Then add the two results together.

Solution

For the first part, $V_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ for this system becomes $V_b-V_a=\text{−}{\displaystyle {\int }_a^b\frac{k_{e}q}{r^2}\widehat{\text{r}}·\widehat{\text{r}}dr}$ which computes to

$$\begin{array}{cc}\text{Δ}V& =\text{−}{\displaystyle {\int }_a^b\frac{k_{e}q}{r^2}dr}=k_{e}q\left[\frac{1}{a}-\frac{1}{b}\right]\hfill \\ & =\left(8.99\times {10}^9{\text{Nm}}^2{\text{/C}}^2\right)\left(2.0\times {10}^{\mathrm{-9}}\text{C}\right)\left[\frac{1}{0.040\text{m}}-\frac{1}{0.12\text{m}}\right]=300\text{V}\text{.}\hfill \end{array}$$

For the second step, $V_B-V_A=\text{−}{\displaystyle {\int }_A^B\overrightarrow{\text{E}}·d\overrightarrow{\text{l}}}$ becomes $\text{Δ}V=\text{−}{\displaystyle {\int }_0^{24\text{°}}\frac{k_{e}q}{r^2}\widehat{\text{r}}·r\widehat{\phi }d\phi }$, but $\widehat{\text{r}}·\widehat{\phi }=0$ and therefore $\text{Δ}V=0.$ Adding the two parts together, we get 300 V.

Significance

We have demonstrated the use of the integral form of the potential difference to obtain a numerical result. Notice that, in this particular system, we could have also used the formula for the potential due to a point charge at the two points and simply taken the difference.

Before presenting problems involving electrostatics, we suggest a problem-solving strategy to follow for this topic.