Electric Potential and Electric Field
12 Electric Potential in a Uniform Electric Field
Learning Objectives
- Describe the relationship between voltage and electric field.
- Derive an expression for the electric potential and electric field.
- Calculate electric field strength given distance and voltage.
In the previous section, we examined the relationship between voltage and energy. We now connect voltage to something more fundamental: the electric field. This connection is extremely important conceptually. Voltage is most closely related to energy, while the electric field is most closely related to force. Understanding how they are connected allows us to move freely between energy-based reasoning and force-based reasoning.
A useful physical example is a uniform electric field produced between two large parallel metal plates, as shown in Figure 12.1. One plate is positively charged and the other negatively charged. The potential difference between the plates creates a uniform electric field between them (except near the edges).

Let us carefully derive the relationship between voltage and electric field in this simple but important case.
Suppose a positive test charge [latex]q[/latex] moves from plate A (higher potential) to plate B (lower potential). The electric field does work on the charge. From the definition of potential energy,
Since [latex]\Delta V = V_B - V_A[/latex], we can write the potential difference between the plates in magnitude form as
The work done by the field becomes
Now let us compute the same work using force. In a uniform electric field,
If the charge moves a distance [latex]d[/latex] in the direction of the field, then
Since both expressions describe the same work,
The charge cancels, leaving the fundamental relationship:
or equivalently,
Voltage and Electric Field (Uniform Case)
Valid only when the electric field is uniform and the motion is parallel to the field.
This result is extremely important. It tells us that the electric field is the rate at which voltage changes with distance in space. A large voltage applied across a small distance produces a strong electric field. A small voltage across a large distance produces a weak electric field.
The units also confirm this relationship. Since
the unit of electric field can be written as volts per meter (V/m). We already know electric field can be measured in newtons per coulomb (N/C). Therefore,
These two units describe the same physical quantity.
Physical Interpretation
Conceptually, the electric field tells us how rapidly electric potential changes in space. If voltage changes very quickly over a small distance, the field is strong. If voltage changes slowly over a large distance, the field is weak.
This idea is deeply relevant in biology and medicine. For example:
- Cell membranes maintain voltage differences (~70 mV) across extremely small thicknesses (~5 nm), creating very strong electric fields inside membranes.
- Defibrillators apply large voltages across the chest to create electric fields strong enough to reset cardiac electrical activity.
- Neural stimulation devices rely on controlled electric fields to influence ion movement.
Even relatively small voltages can create large electric fields if applied across very small distances.
Example: Calculating Electric Field from Voltage
Suppose two parallel plates are separated by 2.0 mm and a potential difference of 500 V is applied. What is the electric field strength between the plates?
First convert distance to meters:
Using
This is a very strong electric field. It illustrates how high voltages over small distances can produce intense electric forces on charges.
In summary, voltage and electric field are two complementary ways to describe the same physical situation. Voltage relates to energy per unit charge. Electric field relates to force per unit charge. In a uniform field, they are directly connected through
Understanding this connection allows us to translate between energy-based reasoning and force-based reasoning in electrical systems.
Example 12.1 What Is the Highest Voltage Possible between Two Plates?
Dry air can support a maximum electric field strength of approximately
If the electric field exceeds this value, air begins to ionize. That is, molecules in the air are stripped of electrons, creating free charges. Once free charges are present, the air is no longer an insulator—it becomes a conductor. This allows a spark (electrical discharge) to occur, which reduces the electric field. This phenomenon is called dielectric breakdown.
Suppose two parallel conducting plates are separated by 2.5 cm of dry air. What is the maximum voltage that can exist between them before breakdown occurs?
Strategy
We are given the maximum electric field strength and the plate separation. Since the electric field between parallel plates is uniform, we use
Solution
First convert the separation distance to meters:
Now multiply by the maximum field:
Discussion
This means that approximately 75 kV is required to produce a spark across a 2.5 cm air gap under dry conditions. If the distance were doubled to 5.0 cm, about 150 kV would be required.
In practice, breakdown often occurs at lower voltages if the conductor surfaces are sharp or pointed. Sharp points concentrate electric field lines, increasing the local field strength. Humid air also breaks down at lower fields because water molecules facilitate ion formation. This is why static sparks are more common on dry winter days.
Understanding dielectric breakdown is important in medical device design, high-voltage equipment, and defibrillation systems, where unintended sparking must be avoided.

Example 12.2 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 carrying a charge of 0.500 μC placed between the plates?
Strategy
When a single electron gains 25.0 keV of energy, it has been accelerated through a potential difference of 25.0 kV. Using
we can determine the electric field. Then we apply
to determine the force.
Solution (a)
Convert distance to meters:
Since 25.0 keV corresponds to 25.0 kV:
Solution (b)
Discussion
The units are newtons because
Since the electric field between parallel plates is uniform, the force is the same everywhere between the plates. This principle is essential in electron microscopes, X-ray tubes, and medical imaging systems, where electrons are accelerated through known voltages.
So far we have considered the special case of a uniform electric field. In more general situations, the electric field varies from point to point. Even then, an important relationship holds:
This equation tells us that the electric field equals the rate at which voltage changes with distance. The minus sign indicates that the electric field points in the direction of decreasing potential. A positive charge naturally moves “downhill” in potential.
Relationship between Voltage and Electric Field
The electric field is the spatial rate of decrease of electric potential. Large changes in voltage over small distances create strong electric fields.
In continuously varying fields, calculus replaces finite differences, and the electric field is defined as the negative gradient of the electric potential. Conceptually, however, the meaning remains the same: electric fields measure how rapidly voltage changes in space.
Section Summary
- For a uniform electric field between parallel plates:
[latex]V_{AB}=Ed \quad \text{and} \quad E=\frac{V_{AB}}{d}[/latex]
- In general:
[latex]E=-\frac{\Delta V}{\Delta s}[/latex]
The electric field equals the negative rate of change of electric potential with distance.
Conceptual Questions
- Discuss how potential difference and electric field strength are related. Give an example.
- What is the strength of the electric field in a region where the electric potential is constant?
- Will a negative charge, initially at rest, move toward higher or lower potential? Explain why.
Problems & Exercises
- Show that units of V/m and N/C for electric field strength are indeed equivalent.
- What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of [latex]1\text{.}\text{50}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}V[/latex]?
- The electric field strength between two parallel conducting plates separated by 4.00 cm is [latex]7\text{.}\text{50}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{V/m}[/latex]. (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?
- How far apart are two conducting plates that have an electric field strength of [latex]4\text{.}\text{50}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V/m}[/latex] between them, if their potential difference is 15.0 kV?
- (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ([latex]3.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}[/latex]) if the plates are separated by 2.00 mm and a potential difference of [latex]5.0×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V}[/latex] is applied? (b) How close together can the plates be with this applied voltage?
- The voltage across a membrane forming a cell wall is 80.0 mV and the membrane is 9.00 nm thick. What is the electric field strength? (The value is surprisingly large, but correct. Membranes are discussed in Capacitors and Dielectrics and Nerve Conduction—Electrocardiograms.) You may assume a uniform electric field.
- Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction—Electrocardiograms.) What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m? You may assume a uniform electric field.
- Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?
- Find the maximum potential difference between two parallel conducting plates separated by 0.500 cm of air, given the maximum sustainable electric field strength in air to be [latex]3.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}[/latex].
- A doubly charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?
- An electron is to be accelerated in a uniform electric field having a strength of [latex]2\text{.}\text{00}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}[/latex]. (a) What energy in keV is given to the electron if it is accelerated through 0.400 m? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?
Glossary
- scalar
- physical quantity with magnitude but no direction
- vector
- physical quantity with both magnitude and direction