FIGURE 27.17 shows a section of wire of length $L$ with a potential difference $\text{Δ}V=V_{+}-V_{-}$ between the ends. Perhaps the two ends of the wire are connected to a battery. A potential difference represents separated positive and negative charges, and, as you saw earlier, some of these charges move onto the surface of the wire. The nonuniform charge distribution creates an electric field in the wire, and that electric field is now driving current through the wire by pushing the charge carriers.

FIGURE 27.17 The current $I$ is related to the potential difference $\text{Δ}V$.

We found in Chapter 26 that the field and the potential are closely related to each other, with the field pointing “downhill,” perpendicular to the equipotential surfaces. Thus it should come as no surprise that the current through the wire is related to the potential difference between the ends of the wire.

Recall that the electric field component $E_s$ is related to the potential by $E_s=-dV\text{/}ds$. We’re interested in only the electric field strength $E=\left|E_s\right|$, so the minus sign isn’t relevant. The field strength is constant inside a constant-diameter conductor; thus

Equation 27.19 is an important result: The electric field strength inside a constant-diameter conductor—the field that drives the current forward—is simply the potential difference between the ends of the conductor divided by its length.

Now we can use $E$ to find the current $I$ in the conductor. We found earlier that the current density is $J=\sigma E$, and the current in a wire of cross-section area $A$ is related to the current density by $I=JA$. Thus

where $\rho =1\text{/}\sigma$ is the resistivity.

Combining Equations 27.19 and 27.20, we see that the current is

That is, the current is directly proportional to the potential difference between the ends of a conductor. We can cast Equation 27.21 into a more useful form if we define the resistance of a conductor to be

The resistance is a property of a specific conductor because it depends on the conductor’s length and diameter as well as on the resistivity of the material from which it is made.

The SI unit of resistance is the ohm, defined as

The ohm is the SI unit of resistance, although kilohms $(1\text{k}\text{Ω}={10}^3\text{Ω})$ and megohms $(1\text{M}\text{Ω}={10}^6\text{Ω})$ are widely used. You can now see from Equation 27.22 why the resistivity $\rho$ has units of $\text{Ω}\text{m}$ while the units of conductivity $\sigma$ are ${\text{Ω}}^{-1}\text{m}{}^{-1}$.

The resistance of a wire or conductor increases as the length increases. This seems reasonable because it should be harder to push electrons through a longer wire than a shorter one. Decreasing the cross-section area also increases the resistance. This again seems reasonable because the same electric field can push more electrons through a fat wire than a skinny one.

The definition of resistance allows us to write the current through a conductor as

In other words, establishing a potential difference $\text{Δ}V$ between the ends of a conductor of resistance $R$ creates an electric field (via the nonuniform distribution of charges on the surface) that, in turn, causes a current $I=\text{Δ}V\text{/}R$ through the conductor. The smaller the resistance, the larger the current. This simple relationship between potential difference and current is known as Ohm’s law.

EXAMPLE 27.6 The resistivity of a leaf

Resistivity measurements on the leaves of corn plants are a good way to assess stress and the plant’s overall health. To determine resistivity, the current is measured when a voltage is applied between two electrodes placed 20 cm apart on a leaf that is 2.5 cm wide and 0.20 mm thick. The following data are obtained by using several different voltages:

Voltage (V)	Current $(\mu A)$
5.0	2.3
10.0	5.1
15.0	7.5
20.0	10.3
25.0	12.2

What is the resistivity of the leaf tissue?

MODEL Model the leaf as a bar of length $L=0.20\text{m}$ with a rectangular cross-section area $A=(0.025\text{ m})(2.0\times {10}^{-4}\text{ m})=5.0\times {10}^{-6}{\text{ m}}^2$. The potential difference creates an electric field inside the leaf and causes a current. The current and the potential difference are related by Ohm’s law.

SOLVE We can find the leaf’s resistivity $\rho$ from its resistance $R$. Ohm’s law

$$I={\displaystyle \frac{1}{R}}\text{Δ}V$$

tells us that a graph of current versus potential difference should be a straight line through the origin with slope $1\text{/}R$. The graph of the data in FIGURE 27.18 is as expected. Using the slope of the best-fit line, $0.50\mu \text{A}\text{/}\text{V}$, we find the leaf’s resistance to be

$$R={\displaystyle \frac{1}{0.50\mu \text{A}\text{/}\text{V}}}=2.0\times {10}^6{\displaystyle \frac{\text{V}}{\text{A}}}=2.0\times {10}^6\text{Ω}$$

FIGURE 27.18 A graph of current versus potential difference.

FIGURE 27.18 Full Alternative Text

We can now use Equation 27.22 to find the resistivity:

$$\rho ={\displaystyle \frac{AR}{L}}={\displaystyle \frac{(5.0\times {10}^{-6}{\text{m}}^2)(2.0\times {10}^6\text{Ω})}{0.20\text{m}}}=50\text{Ω}\text{m}$$

REVIEW This is a huge resistivity compared to metals, but that’s not surprising; the conductivity of the salty fluids in a leaf is certainly much less than that of a metal. In fact, this value is typical of the resistivities of plant and animal tissues.