Electric Potential and Electric Field
12 Electric Potential in a Uniform Electric Field
Learning Objectives
- Describe the relationship between voltage and electric field.
- Derive the relationship between electric potential difference and electric field in a uniform electric field.
- Calculate electric field strength from the potential difference and the distance between two points.
Voltage and Electric Field
In the previous section, we introduced electric potential, potential difference, and electric potential energy. We now connect these energy concepts to a quantity we already know well: the electric field.
This relationship is one of the most important ideas in electrostatics. The electric field tells us how strongly a charge is pushed or pulled, while voltage tells us how much electric potential energy is available per unit charge. Because force and energy are fundamentally related, electric field and voltage are also closely related. Understanding this connection allows us to analyze electrical systems using whichever description—forces or energy—is most convenient.
We begin with the simplest situation: a uniform electric field. Such a field is approximately produced between two large, parallel conducting plates carrying equal and opposite charges, as shown in Figure 12.1. Except near the edges, the electric field between the plates has nearly constant magnitude and direction, making it much easier to analyze than more complicated electric fields.
To derive the relationship between electric field and voltage, consider a positive test charge q moving from the positively charged plate (A) to the negatively charged plate (B). As the charge moves, the electric field does work on it, causing its electric potential energy to decrease.
From the definition of electric potential energy, the work done by the electric field is
Since
we can define the positive potential difference between the plates as
so that the work done by the field becomes
We can obtain the same result using the definition of work. In a uniform electric field, the electric force acting on the charge is
If the charge moves a distance d parallel to the electric field, the work done by the field is
Since both expressions represent the same work,
The charge cancels from both sides, leaving the fundamental relationship for a uniform electric field:
or, equivalently,
Voltage and Electric Field in a Uniform Field
Important: These equations apply only when the electric field is uniform and the displacement is parallel to the field.
This relationship has an intuitive physical interpretation. The electric field measures how rapidly the electric potential changes with distance. If a large potential difference occurs over a very short distance, the electric field is strong. If the same potential difference is spread over a much greater distance, the electric field is weaker.
Another way to think about this is that the electric field represents the voltage gradient—the change in voltage per unit distance. Steep voltage gradients correspond to strong electric fields, while gentle voltage gradients correspond to weak electric fields.
The units also reflect this relationship. Since
the electric field can be expressed in volts per meter (V/m). Earlier, we defined electric field as the force per unit charge, giving the units newtons per coulomb (N/C). These units are exactly equivalent:
Although these units come from different definitions—one based on energy and the other on force—they describe the same physical quantity. Depending on the problem, one form may be more convenient than the other.
Physical Interpretation
The relationship
provides a simple physical interpretation of the electric field. Rather than thinking of the electric field only as a force acting on charges, we can also view it as a measure of how rapidly electric potential changes with distance.
If the potential changes by a large amount over a short distance, the electric field is strong. Conversely, if the same potential difference is spread over a much larger distance, the electric field is weaker. In other words, the electric field describes the steepness of the electric potential, much like the slope of a hill describes how quickly its height changes.
This analogy is closely related to gravitational potential energy. On a gentle hillside, the height changes slowly with distance, so an object experiences only a small component of the gravitational force along the slope. On a steep hillside, the height changes much more rapidly, producing a much larger force that accelerates the object downhill. Likewise, when voltage changes rapidly over a short distance, the resulting electric field is strong and exerts a larger force on charged particles.
This concept is especially important in biology and medicine, where large electric fields are often created across extremely small distances:
- Cell membranes maintain potential differences of about 70 mV across membranes only about 5 nm thick, producing electric fields on the order of 107 V/m.
- Defibrillators apply large voltages across the chest to generate electric fields capable of restoring a normal heart rhythm during certain life-threatening arrhythmias.
- Neural stimulation devices, such as deep brain stimulators and peripheral nerve stimulators, create carefully controlled electric fields that influence the movement of ions across cell membranes and modify nerve activity.
These examples illustrate an important principle: even relatively small voltages can produce extremely large electric fields when they are applied across very small distances. This is one reason why electrical phenomena are so important at the cellular and molecular scales.
Example 12.1: Calculating the Electric Field Between Parallel Plates
Two large parallel conducting plates are separated by 2.0 mm and maintained at a potential difference of 500 V. Assuming the electric field between the plates is uniform, determine the magnitude of the electric field.
Strategy
Because the plates are large compared with their separation, the electric field between them is approximately uniform (except near the edges). We can therefore use the relationship
where d is the separation between the plates. Before substituting numerical values, convert the distance to SI units.
Solution
First convert the plate separation to meters:
Now substitute the known values into the equation:
Since
the electric field may also be written as
Discussion
Although the applied voltage is only 500 V, the electric field is very strong because the plates are separated by only 2.0 mm. This illustrates an important principle: a strong electric field can be produced either by applying a large potential difference or by applying a moderate potential difference across a very small distance.
The same principle explains why cell membranes, despite maintaining voltage differences of only a few tens of millivolts, produce electric fields on the order of 107 V/m. Their voltage changes occur across membranes only a few nanometers thick, creating extremely large voltage gradients.
Key Ideas
Electric potential and electric field provide two complementary ways of describing the same physical situation.
- Electric potential (voltage) describes the electric potential energy available per unit charge.
- Electric field describes the electric force experienced per unit charge.
For a uniform electric field, these quantities are directly related by
This equation shows that the electric field is the rate at which electric potential changes with distance. A large change in voltage over a short distance produces a strong electric field, while the same voltage change spread over a greater distance produces a weaker field.
Being able to move between an energy-based description (voltage) and a force-based description (electric field) is one of the most powerful tools in electrostatics. It allows us to analyze electrical systems from whichever perspective is most convenient and forms the foundation for understanding capacitors, electric circuits, and many biological and medical applications.
In the next section, we will extend these ideas to situations in which the electric field is not uniform and the electric potential changes continuously from one point to another.
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Example 12.2: What Is the Highest Voltage Possible Between Two Plates?
Dry air behaves as an electrical insulator only up to a certain electric field strength. Under normal atmospheric conditions, the maximum electric field that dry air can sustain is approximately
If the electric field exceeds this value, air molecules become ionized. Electrons are stripped from atoms, producing free electrons and positive ions. Once enough free charges are present, the air is no longer an insulator—it becomes electrically conducting, allowing a spark (an electrical discharge) to occur. This process is known as dielectric breakdown.
Suppose two large parallel conducting plates are separated by 2.5 cm of dry air. What is the greatest potential difference that can exist between the plates before dielectric breakdown occurs?
Strategy
Because the electric field between large parallel plates is approximately uniform, we use the relationship
where E is the maximum electric field that air can support and d is the separation between the plates.
Solution
First convert the separation distance to SI units:
Now substitute the known values:
or
Discussion
Under ideal dry-air conditions, a potential difference of approximately 75 kV is sufficient to produce a spark across a 2.5 cm air gap. Increasing the plate separation increases the breakdown voltage proportionally because a larger voltage is required to produce the same electric field. For example, doubling the gap to 5.0 cm increases the breakdown voltage to approximately 150 kV.
In practice, dielectric breakdown often occurs at lower voltages than predicted by this simple calculation. Sharp points or edges on conductors concentrate electric field lines, producing much larger local electric fields that ionize the surrounding air first. Humidity, dust, and other impurities can also lower the breakdown voltage by making ionization easier.
Understanding dielectric breakdown is essential in the design of high-voltage equipment, medical devices, and electrical insulation systems. Engineers must ensure that electric fields remain below the breakdown limit whenever unintended sparks could damage equipment, interfere with sensitive electronics, or create safety hazards.
Example 12.3: Field and Force Inside an Electron Gun
An electron gun consists of two conducting plates separated by 4.00 cm. Electrons accelerated between the plates gain 25.0 keV of kinetic energy.
- What is the electric field strength between the plates?
- What force would this field exert on a small piece of plastic carrying a charge of 0.500 μC placed between the plates?
Strategy
When a single electron gains 25.0 keV of kinetic energy, it has been accelerated through a potential difference of 25.0 kV. (Recall that an electron gains 1 eV of energy for every volt through which it is accelerated.) Since the electric field between large parallel plates is approximately uniform, we first calculate the electric field using
Once the electric field is known, we determine the force on the charged object using
Solution (a)
First convert the plate separation to meters:
Since the electron gains 25.0 keV of energy, the potential difference between the plates is
Substituting into the uniform-field equation gives
Solution (b)
The electric force on the charged plastic is
Discussion
The units are consistent because
Since the electric field between parallel plates is approximately uniform, every charged object experiences the same force per unit charge regardless of its location between the plates. This predictable behavior makes uniform electric fields extremely useful for accelerating and steering charged particles.
Electron guns based on this principle are found in devices such as electron microscopes, cathode-ray tubes, particle accelerators, and X-ray tubes used in medical imaging. In each case, carefully controlled electric fields accelerate charged particles to the desired energy before they interact with a target or sample.
Electric Field as the Rate of Change of Voltage
So far, we have considered the special case of a uniform electric field, in which the electric field has the same magnitude and direction everywhere. In that situation, the relationship between electric field and potential difference is simply
Many real electric fields, however, are not uniform. Around isolated charges, for example, both the magnitude and direction of the electric field change continuously from one point to another. As a result, the electric potential no longer changes at a constant rate with distance.
Even in these more general situations, the electric field and electric potential remain closely related. The relationship becomes
This equation states that the electric field equals the rate at which the electric potential changes with distance. The minus sign indicates that the electric field always points in the direction of decreasing electric potential. A positive test charge released in the field naturally moves "downhill" toward lower electric potential, while a negative charge moves in the opposite direction because it experiences a force opposite to the electric field.
Electric Field and Voltage
Key idea: The electric field points in the direction of the greatest decrease in electric potential. Large changes in voltage over short distances produce strong electric fields.
In continuously varying electric fields, calculus provides a more precise description. Instead of using finite differences, the electric field is defined as the negative gradient of the electric potential. Although the mathematics becomes more sophisticated, the physical interpretation remains the same: the electric field measures how rapidly electric potential changes from one location to another. The steeper the change in potential, the stronger the electric field.
Section Summary
- In a uniform electric field, such as the field between two large parallel conducting plates, the electric field and potential difference are related by
[latex]V_{AB}=Ed \qquad \text{and} \qquad E=\frac{V_{AB}}{d}[/latex]
A larger potential difference or a smaller plate separation produces a stronger electric field.
- The electric field can be expressed in either volts per meter (V/m) or newtons per coulomb (N/C):
[latex]1\ \text{V/m}=1\ \text{N/C}[/latex]
These are equivalent units that describe the same physical quantity.
- For electric fields that are not uniform, the relationship between electric field and electric potential becomes
[latex]E=-\frac{\Delta V}{\Delta s}[/latex]
The electric field points in the direction of decreasing electric potential. Large changes in voltage over short distances produce strong electric fields, while small changes over long distances produce weaker fields.
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
- A physical quantity that has magnitude but no direction. Electric potential (voltage), electric potential energy, mass, and temperature are all scalar quantities.
- vector
- A physical quantity that has both magnitude and direction. Electric field, electric force, velocity, and acceleration are all vector quantities.
A physical quantity that has magnitude but no direction. Electric potential (voltage), electric potential energy, mass, and temperature are all scalar quantities.
a physical quantity that has both magnitude and direction
