Voltage between Points A and B

where [latex]{d}[/latex] is the distance from A to B, or the distance between the plates.

Example 1: What Is the Highest Voltage Possible between Two Plates?

Dry air will support a maximum electric field strength of about [latex]{3.0 \times 10^6 \;\text{V} / \text{m}}[/latex]. 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 [latex]{E}[/latex] between the plates and the distance [latex]{d}[/latex] between them. The equation [latex]{V_{\text{AB}} = Ed}[/latex] can thus be used to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

Entering the given values for [latex]{E}[/latex] and [latex]{d}[/latex] gives

or

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

Discussion

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 will cause a spark if there are points on the surface, since points create greater fields 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, say with static electricity, on dry days.

Example 2: Field and Force inside an Electron Gun

(a) An electron gun has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. What is the electric field strength between the plates? (b) What force would this field exert on a piece of plastic with a [latex]{0.500 \;\mu \text{C}}[/latex] charge that gets between the plates?

Strategy

Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression [latex]{E = \frac{V_{\text{AB}}}{d}}[/latex]. Once the electric field strength is known, the force on a charge is found using [latex]{\textbf{F} = q \textbf{E}}[/latex]. Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, [latex]{F = q \; E}[/latex].

Solution for (a)

The expression for the magnitude of the electric field between two uniform metal plates is

Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for [latex]{V_{\text{AB}}}[/latex] and the plate separation of 0.0400 m, we obtain

Solution for (b)

The magnitude of the force on a charge in an electric field is obtained from the equation

Substituting known values gives

Discussion

Note that the units are newtons, since [latex]{ 1 \;\text{V} / \text{m} = 1 \;\textbf{N} / \text{C}}[/latex]. The force on the charge is the same no matter where the charge is located between the plates. This is because the electric field is uniform between the plates.

Relationship between Voltage and Electric Field

In equation form, the general relationship between voltage and electric field is

where [latex]{\Delta s}[/latex] is the distance over which the change in potential, [latex]{ \Delta V}[/latex], takes place. The minus sign tells us that [latex]\textbf{E}[/latex] points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential.