7.2 Electric potential and potential difference  (Page 5/12)

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 ( [link] ). 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.

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 [link] to move a positive charge q from A , the positive plate, higher potential, to B , the negative plate, lower potential, is

The potential difference between points A and B is

Entering this into the expression for work yields

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 [link] . 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 [link] into our definition for the potential difference between points A and B , we obtain