19.2 Electric Potential in a Uniform Electric Field - College Physics 2e | OpenStax (2024)

Table of Contents

Learning Objectives Voltage between Points A and B Example 19.4 What Is the Highest Voltage Possible between Two Plates? Strategy Solution Discussion Example 19.5 Field and Force inside an Electron Gun Strategy Solution for (a) Solution for (b) Discussion Relationship between Voltage and Electric Field FAQs References

Learning Objectives

By the end of this section, you will be able to:

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 explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric field $E$ is produced by placing a potential difference (or voltage) $Δ V$ across two parallel metal plates, labeled A and B. (See Figure 19.5.) Examining this 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 $Δ V$ or $E$ can be used to describe any charge distribution. $Δ V$ is most closely tied to energy, whereas $E$ is most closely related to force. $Δ V$ is a scalar quantity and has no direction, while $E$ is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by $E$ below.) The relationship between $Δ V$ and $E$ is revealed by calculating the work done by the force in moving a charge from point A to point B. But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.

19.2 Electric Potential in a Uniform Electric Field - College Physics 2e | OpenStax (1)

Figure 19.5 The relationship between $V$ and $E$ for parallel conducting plates is $E = V / d$ . (Note that $Δ V = V_{AB}$ in magnitude. For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows: $-Δ V = V_{A} - V_{B} = V_{AB}$ . See the text for details.)

The work done by the electric field in Figure 19.5 to move a positive charge $q$ from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

$W = –Δ PE = - q Δ V .$

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The potential difference between points A and B is

$-Δ V = - (V_{B} - V_{A}) = V_{A} - V_{B} = V_{AB} .$

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Entering this into the expression for work yields

$W = {qV}_{AB} .$

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Work is $W = Fd cos θ$ ; here $cos θ = 1$ , since the path is parallel to the field, and so $W = Fd$ . Since $F = qE$ , we see that $W = qEd$ . Substituting this expression for work into the previous equation gives

$qEd = {qV}_{AB} .$

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Voltage between Points A and B

$\begin{matrix} \begin{matrix} V_{AB} = Ed \\ E = \frac{V_{AB}}{d} \end{matrix}\} (uniform E - field only), \end{matrix}$

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where $d$ is the distance from A to B, or the distance between the plates.

Example 19.4

What Is the Highest Voltage Possible between Two Plates?

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

Solution

The potential difference or voltage between the plates is

$V_{AB} = Ed .$

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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.

19.2 Electric Potential in a Uniform Electric Field - College Physics 2e | OpenStax (2)

Figure 19.6 A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between adjacent plates is not high enough to cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). (credit: Daderot, Wikimedia Commons)

Example 19.5

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 $0.500 μC$ 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 $E = \frac{V_{AB}}{d}$ . Once the electric field strength is known, the force on a charge is found using $F = q E$ . Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, $F = q E$ .

Solution for (a)

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

$E = \frac{V_{AB}}{d} .$

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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 $V_{AB}$ and the plate separation of 0.0400 m, we obtain

$E = \frac{25 . 0 kV}{0 . 0400 m} = 6 . 25 \times 10^{5} V/m .$

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Solution for (b)

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

$F = qE .$

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Substituting known values gives

$F = (0.500 \times 10^{–6} C) (6.25 \times 10^{5} V/m) = 0.313 N .$

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Discussion

Note that the units are newtons, since $1 V/m = 1 N/C$ . 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.

In more general situations, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of $E$ and also in the direction of lower potential $V$ . Furthermore, the magnitude of $E$ equals the rate of decrease of $V$ with distance. The faster $V$ decreases over distance, the greater the electric field. In equation form, the general relationship between voltage and electric field is

$E = - \frac{Δ V}{Δ s},$

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where $Δ s$ is the distance over which the change in potential, $Δ V$ , takes place. The minus sign tells us that $E$ 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.

Relationship between Voltage and Electric Field

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

$E = - \frac{Δ V}{Δ s},$

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For continually changing potentials, $Δ V$ and $Δ s$ become infinitesimals and differential calculus must be employed to determine the electric field.

19.2 Electric Potential in a Uniform Electric Field - College Physics 2e | OpenStax (2024)

FAQs

What is the electric potential in a uniform electric field? ›

A constant electric field can be produced by placing two large flat conducting plates parallel to each other. The electrical potential difference in a uniform electric field is given as V = E d . The work done moving a charge against the field can be found by W = V q .

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What is the formula for the electric potential in a uniform electric field? ›

A uniform electric field is a field in which the value of the field strength remains the same at all points. If an electric field has the same magnitude and same direction everywhere in a given space then the electric field is said to be uniform.

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What is the formula for electric potential energy in a uniform field? ›

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. E= −ΔVΔs, where Δs is the distance over which the change in potential, ΔV , takes place.

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What is the formula for potential difference in parallel plates? ›

The relationship between V and E for parallel conducting plates is E = V/d. (Note that ΔV = V_AB in magnitude. For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows: –ΔV = V_A – V_B = V_AB. See the text for details.)

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What is the potential surface for a uniform electric field? ›

Uniform field means a constant field at every point and it is directed in a fixed direction for all points. Thus equipotential surfaces must be plane surface. The electric field is always perpendicular to equipotential surface because →E=−→∇V.

What is the formula for the potential difference in an electric field? ›

In a uniform electric field, the equation to calculate the electric potential difference is super easy: V = Ed. In this equation, V is the potential difference in volts, E is the electric field strength (in newtons per coulomb), and d is the distance between the two points (in meters).

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What is the formula for the electric force in a uniform electric field? ›

Once the electric field strength is known, the force on a charge is found using F=qE. Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, F=qE.

What is the electric field electric potential equation? ›

In vector calculus notation, the electric field is given by the negative of the gradient of the electric potential, E = −grad V. This expression specifies how the electric field is calculated at a given point.

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What is the formula for electric potential eV? ›

An electron volt is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form, 1 eV=(1.60×10-19C)(1V)=(1.60×10-19C)(1J/C)=1.60×10-19J. 1 eV = ( 1.60 × 10 -19 C ) ( 1 V ) = ( 1.60 × 10 -19 C ) ( 1 J / C ) = 1.60 × 10 -19 J.

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What is the potential energy of a charge in a uniform field? ›

Here we're free to define the zero for potential energy at some convenient level. If a positive charge moves opposite to the field, the change in potential energy is positive. If it moves in the direction of the field ΔU_e is negative.

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Is potential constant in a uniform electric field? ›

Related to your specific question, if the electric field is constant, then the slope of the potential is a constant which means that the potential is changing linearly. If the potential is constant, then the slope of the potential is zero, which means the electric field is zero.

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What is the formula for the electric field? ›

What is electric field formula? The electric field formula for a charge Q at a point a distance of r from it is written as E = (kQ)/(r^2). The electric field formula gives its strength, sometimes referred to as the magnitude of the electric field.

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Is the electric potential the same everywhere in a uniform field? ›

When we have to move a charge against the electric field we have to do work, against the electric field. And this work done depends on the displacement. So even if the electric field is constant the displacemet is different. So the product (potential) will be different at different point at different distances.

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What is the electric potential of the electric field? ›

Basically, given an electric field, the first step in finding the electrical potential is to pick a point →x0 to have V(→x0)=0. Then, to determine the potential at any point →x, you integrate →E⋅d→s along any path from →x0 to →x.

What is electric potential at a point in electric field? ›

The electric potential at a point is the amount of work done to move a unit positive charge from an infinitely long distance to that point. The SI unit of electric potential is volt (V), and it can also be written as Joule per Coulomb.

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