Electric Field And Voltage Relationship

19 Electric Potential and Electric Field

146 19.ii Electric Potential in a Uniform Electric Field

Learning Objectives

Describe the relationship between voltage and electrical field.
Derive an expression for the electrical potential and electric field.
Calculate electric field strength given distance and voltage.

In the previous section, we explored the human relationship betwixt voltage and energy. In this department, nosotros will explore the relationship between voltage and electric field. For instance, a uniform electrical field [latex]\textbf{Eastward}[/latex] is produced by placing a potential difference (or voltage) [latex]\boldsymbol{\Delta 5}[/latex] across two parallel metallic plates, labeled A and B. (See Effigy ane.) Examining this will tell the states what voltage is needed to produce a certain electric field strength; information technology will also reveal a more fundamental relationship between electric potential and electric field. From a physicist'southward betoken of view, either [latex]\boldsymbol{ \Delta V}[/latex] or [latex]\textbf{Due east}[/latex] can be used to describe any accuse distribution. [latex]\boldsymbol{ \Delta 5}[/latex] is nearly closely tied to energy, whereas [latex]\textbf{E}[/latex] is most closely related to force. [latex]\boldsymbol{\Delta V}[/latex] is a scalar quantity and has no management, while [latex]\textbf{Due east}[/latex] is a vector quantity, having both magnitude and management. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by [latex]\textbf{E}[/latex] below.) The relationship between [latex]\boldsymbol{\Delta V}[/latex] and [latex]\textbf{Eastward}[/latex] is revealed past calculating the work done by the forcefulness in moving a charge from bespeak A to point B. Only, as noted in Affiliate xix.ane Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electrical field as an interesting special instance.

The figure shows two vertically oriented parallel plates A and B separated by a distance d. The plate A is positively charged and B is negatively charged. Electric field lines are parallel between the plates and curved at the ends of the plates. A charge q is moved from A to B. The work done W equals q times V sub A B, and the electric field intensity E equals V sub A B over d and potential difference delta V equals q times V sub A B. — **Figure 1.** The relationship between *Five* and *East* for parallel conducting plates is E = 5/d. (Note that ΔV = V _AB in magnitude. For a accuse that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to exist included as follows: –ΔV = V _A– V _B = V _AB. Encounter the text for details.)

The piece of work done by the electric field in Figure 1 to move a positive charge [latex]\boldsymbol{q}[/latex] from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

[latex]\boldsymbol{Due west = -\Delta \textbf{PE} = -q \Delta V.}[/latex]

The potential difference between points A and B is

[latex]\boldsymbol{- \Delta V = -(V_{\textbf{B}} - V_{\textbf{A}}) = V_{\textbf{A} - V_{\textbf{B}}} = V_{\textbf{AB}}}[/latex].

Entering this into the expression for work yields

[latex]\boldsymbol{W = qV_{\textbf{AB}}}[/latex].

Work is [latex]\boldsymbol{Westward = Fd \;\textbf{cos} \theta}[/latex], since the path is parallel to the field, then [latex]\boldsymbol{W = Fd}[/latex]. Since [latex]\boldsymbol{F = qE}[/latex], we encounter that [latex]\boldsymbol{W = qEd}[/latex]. Substituting this expression for piece of work into the previous equation gives

[latex]\boldsymbol{qEd = qV_{\textbf{AB}}}[/latex].

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

[latex]\brainstorm{array}{l} \boldsymbol{V_{\textbf{AB}} = Ed} \\ \boldsymbol{E = \frac{V_{\textbf{AB}}}{d}} \end{array}[/latex] [latex]\}[/latex] [latex]\boldsymbol{\textbf{(uniform} \; E \;\textbf{- field merely)}} ,[/latex]

where [latex]\boldsymbol{d}[/latex] is the altitude from A to B, or the distance between the plates in Figure 1. Annotation that the above equation implies the units for electrical 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:

[latex]\boldsymbol{1 \;\textbf{North} / \textbf{C} = 1 \;\textbf{V} / \textbf{yard}}.[/latex]

Voltage between Points A and B

[latex]\begin{array}{l} \boldsymbol{V_{\textbf{AB}} = Ed} \\ \boldsymbol{East = \frac{V_{\textbf{AB}}}{d}} \end{assortment}[/latex] [latex]\}[/latex] [latex]\boldsymbol{\textbf{(uniform} \; East \;\textbf{- field only)}} ,[/latex]

where [latex]\boldsymbol{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 electrical field strength of almost [latex]\boldsymbol{three.0 \times x^6 \;\textbf{Five} / \textbf{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 ii parallel conducting plates separated by 2.v cm of dry air?

Strategy

We are given the maximum electric field [latex]\boldsymbol{E}[/latex] betwixt the plates and the distance [latex]\boldsymbol{d}[/latex] between them. The equation [latex]\boldsymbol{V_{\textbf{AB}} = Ed}[/latex] can thus be used to calculate the maximum voltage.

Solution

The potential difference or voltage between the plates is

[latex]\boldsymbol{V_{\textbf{AB}} = Ed}.[/latex]

Entering the given values for [latex]\boldsymbol{Eastward}[/latex] and [latex]\boldsymbol{d}[/latex] gives

[latex]\boldsymbol{V_{\textbf{AB}} = (3.0 \times 10^6 \;\textbf{Five} / \textbf{chiliad})(0.025 \;\textbf{m}) = 7.5 \times 10^4 \;\textbf{V}}[/latex]

[latex]\boldsymbol{V_{\textbf{AB}} = 75 \;\textbf{kV}} .[/latex]

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

Give-and-take

One of the implications of this result is that information technology takes almost 75 kV to brand a spark jump beyond 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 downwards at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built upward, say with static electricity, on dry days.

The picture shows a spark chamber placed on a wooden base. — **Figure ii.** A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas betwixt the plates allows a spark to spring. The sparks are perpendicular to the plates, following electrical field lines between them. The potential divergence betwixt adjacent plates is non loftier plenty to cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). (credit: Daderot, Wikimedia Commons)

Instance 2: Field and Forcefulness within an Electron Gun

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

Strategy

Since the voltage and plate separation are given, the electric field strength can be calculated straight from the expression [latex]\boldsymbol{East = \frac{V_{\textbf{AB}}}{d}}[/latex]. Once the electric field strength is known, the strength on a accuse is found using [latex]\boldsymbol{\textbf{F} = q \textbf{Due east}}[/latex]. Since the electric field is in but one direction, we can write this equation in terms of the magnitudes, [latex]\boldsymbol{F = q \; Due east}[/latex].

Solution for (a)

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

[latex]\boldsymbol{E =}[/latex] [latex]\boldsymbol{ \frac{V_{\textbf{AB}}}{d}}[/latex].

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

[latex]\boldsymbol{E=}[/latex] [latex]\boldsymbol{\frac{25.0 \;\textbf{kV}}{0.0400 \;\textbf{thou}}}[/latex] [latex]\boldsymbol{= half dozen.25 \times x^5 \;\textbf{V} / \textbf{m}}.[/latex]

Solution for (b)

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

[latex]\boldsymbol{F=qE}.[/latex]

Substituting known values gives

[latex]\boldsymbol{F = (0.500 \times ten^{-6} \;\textbf{C})(vi.25 \times x^5 \;\textbf{V} / \textbf{m}) = 0.313 \;\textbf{N}}.[/latex]

Discussion

Note that the units are newtons, since [latex]\boldsymbol{ ane \;\textbf{V} / \textbf{m} = 1 \;\textbf{Northward} / \textbf{C}}[/latex]. The force on the charge is the same no matter where the accuse is located between the plates. This is considering the electric field is compatible between the plates.

In more general situations, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, considering the force on a positive accuse is in the direction of [latex]\textbf{Eastward}[/latex] and too in the management of lower potential [latex]\boldsymbol{V}[/latex]. Furthermore, the magnitude of [latex]\textbf{E}[/latex] equals the rate of decrease of [latex]\boldsymbol{V}[/latex] with altitude. The faster [latex]\boldsymbol{V}[/latex] decreases over distance, the greater the electrical field. In equation form, the general human relationship between voltage and electric field is

[latex]\boldsymbol{East =}[/latex] [latex]\boldsymbol{-\frac{\Delta V}{\Delta s}} ,[/latex]

where [latex]\boldsymbol{ \Delta s}[/latex] is the altitude over which the modify in potential, [latex]\boldsymbol{\Delta Five}[/latex], takes place. The minus sign tells us that [latex]\textbf{Eastward}[/latex] points in the direction of decreasing potential. The electric field is said to be the gradient (every bit in class or slope) of the electric potential.

Relationship between Voltage and Electric Field

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

[latex]\boldsymbol{E =}[/latex] [latex]\boldsymbol{- \frac{\Delta Five}{\Delta s}} ,[/latex]

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

For continually irresolute potentials, [latex]\boldsymbol{\Delta V}[/latex] and [latex]\boldsymbol{\Delta s}[/latex] become infinitesimals and differential calculus must exist employed to determine the electric field.

Section Summary

The voltage between points A and B is
[latex]\brainstorm{array}{l} \boldsymbol{V_{\textbf{AB}} = Ed} \\ \boldsymbol{E = \frac{V_{\textbf{AB}}}{d}} \end{array}[/latex] [latex]\}[/latex] [latex]\boldsymbol{\textbf{(uniform} \; E \;\textbf{- field merely)}} ,[/latex]

where [latex]\boldsymbol{d}[/latex] is the distance from A to B, or the distance betwixt the plates.
In equation form, the general human relationship betwixt voltage and electrical field is
[latex]\boldsymbol{Due east =}[/latex] [latex]\boldsymbol{- \frac{\Delta V}{\Delta due south}} ,[/latex]

where [latex]\boldsymbol{\Delta due south}[/latex] is the distance over which the modify in potential, [latex]\boldsymbol{\Delta 5}[/latex], takes identify. The minus sign tells us that [latex]\textbf{East}[/latex] points in the direction of decreasing potential.) The electric field is said to be the slope (every bit in grade or gradient) of the electrical potential.

Conceptual Questions

one: Discuss how potential deviation and electrical field force are related. Give an example.

2: What is the strength of the electric field in a region where the electric potential is constant?

three: Will a negative charge, initially at residue, motion toward college or lower potential? Explain why.

Problems & Exercises

ane: Show that units of 5/chiliad and Northward/C for electric field strength are indeed equivalent.

two: What is the strength of the electric field betwixt two parallel conducting plates separated by ane.00 cm and having a potential difference (voltage) between them of [latex]\boldsymbol{one.50 \times ten^iv \;\textbf{Five}}[/latex]?

iii: The electric field forcefulness between 2 parallel conducting plates separated by 4.00 cm is [latex]\boldsymbol{vii.50 \times 10^four \;\textbf{V} / \textbf{thou}}[/latex]. (a) What is the potential departure 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 iii.00 cm from the other)?

iv: How far apart are two conducting plates that have an electric field strength of [latex]\boldsymbol{four.50 \times ten^3 \;\textbf{V} / \textbf{k}}[/latex] between them, if their potential difference is 15.0 kV?

5: (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air [latex]\boldsymbol{(iii.0 \times x^6 \;\textbf{5} / \textbf{m})}[/latex] if the plates are separated by two.00 mm and a potential divergence of [latex]\boldsymbol{5.0 \times 10^3 \;\textbf{Five}}[/latex] is applied? (b) How close together can the plates exist with this applied voltage?

6: The voltage across a membrane forming a cell wall is 80.0 mV and the membrane is 9.00 nm thick. What is the electrical field force? (The value is surprisingly large, merely correct. Membranes are discussed in Chapter 19.5 Capacitors and Dielectrics and Chapter 20.seven Nerve Conduction—Electrocardiograms.) Y'all may presume a uniform electric field.

vii: Membrane walls of living cells have surprisingly big electric fields across them due to separation of ions. (Membranes are discussed in some detail in Affiliate 20.vii Nerve Conduction—Electrocardiograms.) What is the voltage across an viii.00 nm–thick membrane if the electric field strength beyond information technology is v.l MV/m? Yous may presume a uniform electrical field.

viii: Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to exist at cypher volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the cipher volt plate (and 2.00 cm from the other) is 450 Five? (b) What is the voltage between the plates?

nine: 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 exist [latex]\boldsymbol{iii.0 \times ten^6 \;\textbf{V} / \textbf{g}}[/latex].

10: 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 betwixt the plates?

11: An electron is to be accelerated in a compatible electrical field having a strength of [latex]\boldsymbol{ii.00 \times 10^6 \;\textbf{V} / \textbf{m}}[/latex]. (a) What energy in keV is given to the electron if it is accelerated through 0.400 chiliad? (b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV?

Glossary

scalar: concrete quantity with magnitude simply no management

vector: physical quantity with both magnitude and management

Solutions

Problems & Exercises

iii: (a) 3.00 kV

(b) 750 V

five: (a) No. The electrical field strength between the plates is [latex]\boldsymbol{2.5 \times ten^6 \;\textbf{5} / \textbf{thousand}}[/latex], which is lower than the breakdown strength for air [latex]\boldsymbol{(3.0 \times 10^6 \;\textbf{V} / \textbf{chiliad})}[/latex].

(b) ane.vii mm

7: 44.0 mV

9: xv kV

eleven: (a) 800 KeV

(b) 25.0 km