College Physics for AP® Courses

# 19.2Electric Potential in a Uniform Electric Field

College Physics for AP® Courses19.2 Electric Potential in a Uniform Electric Field

## 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.

The information presented in this section supports the following AP® learning objectives and science practices:

• 2.C.5.2 The student is able to calculate the magnitude and determine the direction of the electric field between two electrically charged parallel plates, given the charge of each plate, or the electric potential difference and plate separation. (S.P. 2.2)
• 2.C.5.3 The student is able to represent the motion of an electrically charged particle in the uniform field between two oppositely charged plates and express the connection of this motion to projectile motion of an object with mass in the Earth’s gravitational field. (S.P. 1.1, 2.2, 7.1)
• 2.E.3.1 The student is able to apply mathematical routines to calculate the average value of the magnitude of the electric field in a region from a description of the electric potential in that region using the displacement along the line on which the difference in potential is evaluated. (S.P. 2.2)
• 2.E.3.2 The student is able to apply the concept of the isoline representation of electric potential for a given electric charge distribution to predict the average value of the electric field in the region. (S.P. 1.4, 6.4)

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 $EE size 12{E} {}$ is produced by placing a potential difference (or voltage) $ΔVΔV size 12{V} {}$ across two parallel metal plates, labeled A and B. (See Figure 19.7.) 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 $ΔVΔV size 12{V} {}$ or $EE size 12{E} {}$ can be used to describe any charge distribution. $ΔVΔV size 12{V} {}$ is most closely tied to energy, whereas $EE size 12{E} {}$ is most closely related to force. $ΔVΔV size 12{V} {}$ is a scalar quantity and has no direction, while $EE size 12{E} {}$ is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by $EE size 12{V} {}$ below.) The relationship between $ΔVΔV size 12{V} {}$ and $EE size 12{E} {}$ 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.

Figure 19.7 The relationship between $VV size 12{V} {}$ and $EE size 12{E} {}$ for parallel conducting plates is $E=V/dE=V/d size 12{E=V/d} {}$. (Note that $Δ V = V AB Δ V = V AB size 12{ΔV=V rSub { size 8{"AB"} } } {}$ 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: $–Δ V = V A – V B = V AB –Δ V = V A – V B = V AB$. See the text for details.)

The work done by the electric field in Figure 19.7 to move a positive charge $q q size 12{q} {}$ from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

$W=–ΔPE=–qΔV.W=–ΔPE=–qΔV. size 12{W= - Δ"PE"= - qΔV} {}$
19.21

The potential difference between points A and B is

$–Δ V = – ( V B – V A ) = V A – V B = V AB . –Δ V = – ( V B – V A ) = V A – V B = V AB .$
19.22

Entering this into the expression for work yields

$W = qV AB . W = qV AB . size 12{W= ital "qV" rSub { size 8{ ital "AB"} } } {}$
19.23

Work is $W=FdcosθW=Fdcosθ size 12{W= ital "Fd""cos"?} {}$; here $cosθ=1cosθ=1$, since the path is parallel to the field, and so $W=FdW=Fd$. Since $F=qEF=qE$, we see that $W=qEdW=qEd$. Substituting this expression for work into the previous equation gives

$qEd=qVAB.qEd=qVAB. size 12{qEd= ital "qV" rSub { size 8{ ital "AB"} } } {}$
19.24

The charge cancels, and so the voltage between points A and B is seen to be

$V AB = Ed E = V AB d (uniform E - field only), V AB = Ed E = V AB d (uniform E - field only),$
19.25

where $dd size 12{d} {}$ is the distance from A to B, or the distance between the plates in Figure 19.7. Note that the above equation implies the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid:

$1 N/C=1 V/m.1 N/C=1 V/m. size 12{"1 N"/C="1 V"/m} {}$
19.26

## Voltage between Points A and B

$V AB = Ed E = V AB d (uniform E - field only), V AB = Ed E = V AB d (uniform E - field only),$
19.27

where $dd size 12{d} {}$ 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 $3.0×106 V/m3.0×106 V/m size 12{3×"10" rSup { size 8{6} } " V/m"} {}$. 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 $EE size 12{E} {}$ between the plates and the distance $d d$ between them. The equation $VAB=EdVAB=Ed size 12{V rSub { size 8{"AB"} } =Ed} {}$ can thus be used to calculate the maximum voltage.

### Solution

The potential difference or voltage between the plates is

$VAB=Ed.VAB=Ed. size 12{V rSub { size 8{"AB"} } =Ed} {}$
19.28

Entering the given values for $EE size 12{E} {}$ and $dd size 12{d} {}$ gives

$V AB = ( 3.0 × 10 6 V/m ) ( 0.025 m ) = 7.5 × 10 4 V V AB = ( 3.0 × 10 6 V/m ) ( 0.025 m ) = 7.5 × 10 4 V$
19.29

or

$VAB=75 kV.VAB=75 kV. size 12{V rSub { size 8{"AB"} } ="75" "kV"} {}$
19.30

(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.

Figure 19.8 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)

## Making Connections: Uniform Fields

Figure 19.9 (a) A massive particle launched horizontally in a downward gravitational field will fall to the ground. (b) A positively charged particle launched horizontally in a downward electric field will fall toward the negative potential; a negatively charged particle will move in the opposite direction.

Recall from Projectile Motion (3.4 Projectile Motion) that a massive projectile launched horizontally (for example, from a cliff) in a uniform downward gravitational field (as we find near the surface of the Earth) will follow a parabolic trajectory downward until it hits the ground, as shown in Figure 19.9(a).

An identical outcome occurs for a positively charged particle in a uniform electric field (Figure 19.9(b)); it follows the electric field “downhill” until it runs into something. The difference between the two cases is that the gravitational force is always attractive; the electric force has two kinds of charges, and therefore may be either attractive or repulsive. Therefore, a negatively charged particle launched into the same field will fall “uphill”.

## 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 $0.500 μC 0.500 μC$ 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 $E=VABdE=VABd size 12{E= { {V rSub { size 8{"AB"} } } over {d} } } {}$. Once the electric field strength is known, the force on a charge is found using $F=qEF=qE$. Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, $F=qEF=qE size 12{F=qE} {}$.

### Solution for (a)

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

$E=VABd.E=VABd. size 12{E= { {V rSub { size 8{"AB"} } } over {d} } } {}$
19.31

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 $VABVAB size 12{V rSub { size 8{"AB"} } } {}$ and the plate separation of 0.0400 m, we obtain

$E=25.0 kV0.0400 m=6.25×105 V/m.E=25.0 kV0.0400 m=6.25×105 V/m. size 12{E= { {"25" "." "0 kV"} over {0 "." "0400 m"} } =6 "." "25"´"10" rSup { size 8{5} } " V"/m} {}$
19.32

### Solution for (b)

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

$F=qE.F=qE. size 12{F=qE} {}$
19.33

Substituting known values gives

$F=(0.500×10–6 C)(6.25×105 V/m)=0.313 N.F=(0.500×10–6 C)(6.25×105 V/m)=0.313 N. size 12{F= $$0 "." "500"´"10" rSup { size 8{-6} } " C"$$ $$6 "." "25"´"10" rSup { size 8{5} } " V"/m$$ =0 "." "310"" N"} {}$
19.34

### Discussion

Note that the units are newtons, since $1 V/m=1 N/C1 V/m=1 N/C size 12{1" V/m"=1" N/C"} {}$. 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 $EE$ and also in the direction of lower potential $VV$. Furthermore, the magnitude of $EE$ equals the rate of decrease of $VV size 12{V} {}$ with distance. The faster $VV size 12{V} {}$ decreases over distance, the greater the electric field. In equation form, the general relationship between voltage and electric field is

$E=–ΔVΔs,E=–ΔVΔs, size 12{E= - { {ΔV} over {Δs} } } {}$
19.35

where $ΔsΔs size 12{?s} {}$ is the distance over which the change in potential, $ΔVΔV size 12{ΔV} {}$, takes place. The minus sign tells us that $EE size 12{E} {}$ 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

$E=–ΔVΔs,E=–ΔVΔs, size 12{E= - { {ΔV} over {Δs} } } {}$
19.36

where $ΔsΔs size 12{?s} {}$ is the distance over which the change in potential, $ΔVΔV size 12{?V} {}$, takes place. The minus sign tells us that $EE$ 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.

Note that Equation (19.36) is defining the average electric field over the given region.

For continually changing potentials, $ΔVΔV size 12{ΔV} {}$ and $ΔsΔs size 12{Δs} {}$ become infinitesimals and differential calculus must be employed to determine the electric field.

## Making Connections: Non-Parallel Conducting Plates

Consider two conducting plates, placed as shown in Figure 19.10. If the plates are held at a fixed potential difference ΔV, the average electric field is strongest between the near edges of the plates, and weakest between the two far edges of the plates.

Figure 19.10 Two nonparallel plates, held at a fixed potential difference.

Now assume that the potential difference is 60 V. If the arc length along the field line labeled by A is 10 cm, what is the electric field at point A? How about point B, if the arc length along that field line is 17 cm? How would the density of the electric potential isolines, if they were drawn on the figure, compare at these two points? Can you use this concept to estimate what the electric field strength would be at a point midway between A and B?

Applying Equation. 19.36, we find that the electric fields at A and B are 600 V/m and 350 V/m respectively. The isolines would be denser at A than at B, and would spread out evenly from A to B. Therefore, the electric field at a point halfway between the two would have an arc length of 13.5 cm and be approximately 440 V/m.

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