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University Physics Volume 2

5.4 Electric Field

University Physics Volume 25.4 Electric Field
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  1. Preface
  2. Unit 1. Thermodynamics
    1. 1 Temperature and Heat
      1. Introduction
      2. 1.1 Temperature and Thermal Equilibrium
      3. 1.2 Thermometers and Temperature Scales
      4. 1.3 Thermal Expansion
      5. 1.4 Heat Transfer, Specific Heat, and Calorimetry
      6. 1.5 Phase Changes
      7. 1.6 Mechanisms of Heat Transfer
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 The Kinetic Theory of Gases
      1. Introduction
      2. 2.1 Molecular Model of an Ideal Gas
      3. 2.2 Pressure, Temperature, and RMS Speed
      4. 2.3 Heat Capacity and Equipartition of Energy
      5. 2.4 Distribution of Molecular Speeds
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 The First Law of Thermodynamics
      1. Introduction
      2. 3.1 Thermodynamic Systems
      3. 3.2 Work, Heat, and Internal Energy
      4. 3.3 First Law of Thermodynamics
      5. 3.4 Thermodynamic Processes
      6. 3.5 Heat Capacities of an Ideal Gas
      7. 3.6 Adiabatic Processes for an Ideal Gas
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 The Second Law of Thermodynamics
      1. Introduction
      2. 4.1 Reversible and Irreversible Processes
      3. 4.2 Heat Engines
      4. 4.3 Refrigerators and Heat Pumps
      5. 4.4 Statements of the Second Law of Thermodynamics
      6. 4.5 The Carnot Cycle
      7. 4.6 Entropy
      8. 4.7 Entropy on a Microscopic Scale
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Electricity and Magnetism
    1. 5 Electric Charges and Fields
      1. Introduction
      2. 5.1 Electric Charge
      3. 5.2 Conductors, Insulators, and Charging by Induction
      4. 5.3 Coulomb's Law
      5. 5.4 Electric Field
      6. 5.5 Calculating Electric Fields of Charge Distributions
      7. 5.6 Electric Field Lines
      8. 5.7 Electric Dipoles
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    2. 6 Gauss's Law
      1. Introduction
      2. 6.1 Electric Flux
      3. 6.2 Explaining Gauss’s Law
      4. 6.3 Applying Gauss’s Law
      5. 6.4 Conductors in Electrostatic Equilibrium
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 7 Electric Potential
      1. Introduction
      2. 7.1 Electric Potential Energy
      3. 7.2 Electric Potential and Potential Difference
      4. 7.3 Calculations of Electric Potential
      5. 7.4 Determining Field from Potential
      6. 7.5 Equipotential Surfaces and Conductors
      7. 7.6 Applications of Electrostatics
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 8 Capacitance
      1. Introduction
      2. 8.1 Capacitors and Capacitance
      3. 8.2 Capacitors in Series and in Parallel
      4. 8.3 Energy Stored in a Capacitor
      5. 8.4 Capacitor with a Dielectric
      6. 8.5 Molecular Model of a Dielectric
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 9 Current and Resistance
      1. Introduction
      2. 9.1 Electrical Current
      3. 9.2 Model of Conduction in Metals
      4. 9.3 Resistivity and Resistance
      5. 9.4 Ohm's Law
      6. 9.5 Electrical Energy and Power
      7. 9.6 Superconductors
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 10 Direct-Current Circuits
      1. Introduction
      2. 10.1 Electromotive Force
      3. 10.2 Resistors in Series and Parallel
      4. 10.3 Kirchhoff's Rules
      5. 10.4 Electrical Measuring Instruments
      6. 10.5 RC Circuits
      7. 10.6 Household Wiring and Electrical Safety
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 11 Magnetic Forces and Fields
      1. Introduction
      2. 11.1 Magnetism and Its Historical Discoveries
      3. 11.2 Magnetic Fields and Lines
      4. 11.3 Motion of a Charged Particle in a Magnetic Field
      5. 11.4 Magnetic Force on a Current-Carrying Conductor
      6. 11.5 Force and Torque on a Current Loop
      7. 11.6 The Hall Effect
      8. 11.7 Applications of Magnetic Forces and Fields
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 12 Sources of Magnetic Fields
      1. Introduction
      2. 12.1 The Biot-Savart Law
      3. 12.2 Magnetic Field Due to a Thin Straight Wire
      4. 12.3 Magnetic Force between Two Parallel Currents
      5. 12.4 Magnetic Field of a Current Loop
      6. 12.5 Ampère’s Law
      7. 12.6 Solenoids and Toroids
      8. 12.7 Magnetism in Matter
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    9. 13 Electromagnetic Induction
      1. Introduction
      2. 13.1 Faraday’s Law
      3. 13.2 Lenz's Law
      4. 13.3 Motional Emf
      5. 13.4 Induced Electric Fields
      6. 13.5 Eddy Currents
      7. 13.6 Electric Generators and Back Emf
      8. 13.7 Applications of Electromagnetic Induction
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 14 Inductance
      1. Introduction
      2. 14.1 Mutual Inductance
      3. 14.2 Self-Inductance and Inductors
      4. 14.3 Energy in a Magnetic Field
      5. 14.4 RL Circuits
      6. 14.5 Oscillations in an LC Circuit
      7. 14.6 RLC Series Circuits
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 15 Alternating-Current Circuits
      1. Introduction
      2. 15.1 AC Sources
      3. 15.2 Simple AC Circuits
      4. 15.3 RLC Series Circuits with AC
      5. 15.4 Power in an AC Circuit
      6. 15.5 Resonance in an AC Circuit
      7. 15.6 Transformers
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 16 Electromagnetic Waves
      1. Introduction
      2. 16.1 Maxwell’s Equations and Electromagnetic Waves
      3. 16.2 Plane Electromagnetic Waves
      4. 16.3 Energy Carried by Electromagnetic Waves
      5. 16.4 Momentum and Radiation Pressure
      6. 16.5 The Electromagnetic Spectrum
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Explain the purpose of the electric field concept
  • Describe the properties of the electric field
  • Calculate the field of a collection of source charges of either sign

As we showed in the preceding section, the net electric force on a test charge is the vector sum of all the electric forces acting on it, from all of the various source charges, located at their various positions. But what if we use a different test charge, one with a different magnitude, or sign, or both? Or suppose we have a dozen different test charges we wish to try at the same location? We would have to calculate the sum of the forces from scratch. Fortunately, it is possible to define a quantity, called the electric field, which is independent of the test charge. It only depends on the configuration of the source charges, and once found, allows us to calculate the force on any test charge.

Defining a Field

Suppose we have N source charges q1,q2,q3,…,qNq1,q2,q3,…,qN located at positions r1,r2,r3,…,rNr1,r2,r3,…,rN, applying N electrostatic forces on a test charge Q. The net force on Q is (see Equation 5.2)

F=F1+F2+F3++FN=14πε0(Qq1r12r^1+Qq2r22r^2+Qq3r32r^3++QqNr12r^N)=Q[14πε0(q1r12r^1+q2r22r^2+q3r32r^3++qNr12r^N)].F=F1+F2+F3++FN=14πε0(Qq1r12r^1+Qq2r22r^2+Qq3r32r^3++QqNr12r^N)=Q[14πε0(q1r12r^1+q2r22r^2+q3r32r^3++qNr12r^N)].

We can rewrite this as

F=QEF=QE
(5.3)

where

E14πε0(q1r12r^1+q2r22r^2+q3r32r^3++qNr12r^N)E14πε0(q1r12r^1+q2r22r^2+q3r32r^3++qNr12r^N)

or, more compactly,

E(P)14πε0i=1Nqiri2r^i.E(P)14πε0i=1Nqiri2r^i.
(5.4)

This expression is called the electric field at position P=P(x,y,z)P=P(x,y,z) of the N source charges. Here, P is the location of the point in space where you are calculating the field and is relative to the positions riri of the source charges (Figure 5.18). Note that we have to impose a coordinate system to solve actual problems.

Eight source charges are shown as small spheres distributed within an x y z coordinate system. The sources are labeled q sub 1, q sub 2, and so on. Sources 1, 2, 4, 7 and 8 are shaded red and sources 3, 5, and 6 are shaded blue. A test point is also shown and labeled as point P. The electric field vectors due to each source is shown as an arrow at point P, pointing toward point P and labeled with the index of the associated source. Vector E 1 points away from q 1, E 2 away from q 2, E 4 away from q 4, E 7 away from q 7, and E 8 away from q 8. Vector E 3 points toward q 3, vector E 5 toward q 5, and vector E 6 toward q 6.
Figure 5.18 Each of these eight source charges creates its own electric field at every point in space; shown here are the field vectors at an arbitrary point P. Like the electric force, the net electric field obeys the superposition principle.

Notice that the calculation of the electric field makes no reference to the test charge. Thus, the physically useful approach is to calculate the electric field and then use it to calculate the force on some test charge later, if needed. Different test charges experience different forces Equation 5.3, but it is the same electric field Equation 5.4. That being said, recall that there is no fundamental difference between a test charge and a source charge; these are merely convenient labels for the system of interest. Any charge produces an electric field; however, just as Earth’s orbit is not affected by Earth’s own gravity, a charge is not subject to a force due to the electric field it generates. Charges are only subject to forces from the electric fields of other charges.

In this respect, the electric field EE of a point charge is similar to the gravitational field gg of Earth; once we have calculated the gravitational field at some point in space, we can use it any time we want to calculate the resulting force on any mass we choose to place at that point. In fact, this is exactly what we do when we say the gravitational field of Earth (near Earth’s surface) has a value of 9.81m/s2,9.81m/s2, and then we calculate the resulting force (i.e., weight) on different masses. Also, the general expression for calculating gg at arbitrary distances from the center of Earth (i.e., not just near Earth’s surface) is very similar to the expression for EE: g=GMr2r^g=GMr2r^, where G is a proportionality constant, playing the same role for gg as 14πε014πε0 does for EE. The value of gg is calculated once and is then used in an endless number of problems.

To push the analogy further, notice the units of the electric field: From F=QEF=QE, the units of E are newtons per coulomb, N/C, that is, the electric field applies a force on each unit charge. Now notice the units of g: From w=mgw=mg, the units of g are newtons per kilogram, N/kg, that is, the gravitational field applies a force on each unit mass. We could say that the gravitational field of Earth, near Earth’s surface, has a value of 9.81 N/kg.

The Meaning of “Field”

Recall from your studies of gravity that the word “field” in this context has a precise meaning. A field, in physics, is a physical quantity whose value depends on (is a function of) position, relative to the source of the field. In the case of the electric field, Equation 5.4 shows that the value of EE (both the magnitude and the direction) depends on where in space the point P is located, measured from the locations riri of the source charges qiqi.

In addition, since the electric field is a vector quantity, the electric field is referred to as a vector field. (The gravitational field is also a vector field.) In contrast, a field that has only a magnitude at every point is a scalar field. The temperature in a room is an example of a scalar field. It is a field because the temperature, in general, is different at different locations in the room, and it is a scalar field because temperature is a scalar quantity.

Also, as you did with the gravitational field of an object with mass, you should picture the electric field of a charge-bearing object (the source charge) as a continuous, immaterial substance that surrounds the source charge, filling all of space—in principle, to ±± in all directions. The field exists at every physical point in space. To put it another way, the electric charge on an object alters the space around the charged object in such a way that all other electrically charged objects in space experience an electric force as a result of being in that field. The electric field, then, is the mechanism by which the electric properties of the source charge are transmitted to and through the rest of the universe. (Again, the range of the electric force is infinite.)

We will see in subsequent chapters that the speed at which electrical phenomena travel is the same as the speed of light. There is a deep connection between the electric field and light.

Superposition

Yet another experimental fact about the field is that it obeys the superposition principle. In this context, that means that we can (in principle) calculate the total electric field of many source charges by calculating the electric field of only q1q1 at position P, then calculate the field of q2q2 at P, while—and this is the crucial idea—ignoring the field of, and indeed even the existence of, q1.q1. We can repeat this process, calculating the field of each individual source charge, independently of the existence of any of the other charges. The total electric field, then, is the vector sum of all these fields. That, in essence, is what Equation 5.4 says.

In the next section, we describe how to determine the shape of an electric field of a source charge distribution and how to sketch it.

The Direction of the Field

Equation 5.4 enables us to determine the magnitude of the electric field, but we need the direction also. We use the convention that the direction of any electric field vector is the same as the direction of the electric force vector that the field would apply to a positive test charge placed in that field. Such a charge would be repelled by positive source charges (the force on it would point away from the positive source charge) but attracted to negative charges (the force points toward the negative source).

Direction of the Electric Field

By convention, all electric fields EE point away from positive source charges and point toward negative source charges.

Interactive

Add charges to the Electric Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.

Example 5.3

The E-field of an Atom In an ionized helium atom, the most probable distance between the nucleus and the electron is r=26.5×10−12mr=26.5×10−12m. What is the electric field due to the nucleus at the location of the electron?

Strategy Note that although the electron is mentioned, it is not used in any calculation. The problem asks for an electric field, not a force; hence, there is only one charge involved, and the problem specifically asks for the field due to the nucleus. Thus, the electron is a red herring; only its distance matters. Also, since the distance between the two protons in the nucleus is much, much smaller than the distance of the electron from the nucleus, we can treat the two protons as a single charge +2e (Figure 5.19).

A positive charge of plus 2 e is shown at the center of a sphere of radius r. An electron is depicted as a particle on the sphere. The vector r is shown as a vector with its tail at the center and its head at the location of the electron. The electric field at the location of the electron is shown as a vector E with its tail at the electron and pointing directly away from the center.
Figure 5.19 A schematic representation of a helium atom. Again, helium physically looks nothing like this, but this sort of diagram is helpful for calculating the electric field of the nucleus.

Solution The electric field is calculated by

E=14πε0i=1Nqiri2r^i.E=14πε0i=1Nqiri2r^i.

Since there is only one source charge (the nucleus), this expression simplifies to

E=14πε0qr2r^.E=14πε0qr2r^.

Here q=2e=2(1.6×10−19C)q=2e=2(1.6×10−19C) (since there are two protons) and r is given; substituting gives

E=14π(8.85×10−12C2N·m2)2(1.6×10−19C)(26.5×10−12m)2r^=4.1×1012NCr^.E=14π(8.85×10−12C2N·m2)2(1.6×10−19C)(26.5×10−12m)2r^=4.1×1012NCr^.

The direction of EE is radially away from the nucleus in all directions. Why? Because a positive test charge placed in this field would accelerate radially away from the nucleus (since it is also positively charged), and again, the convention is that the direction of the electric field vector is defined in terms of the direction of the force it would apply to positive test charges.

Example 5.4

The E-Field above Two Equal Charges (a) Find the electric field (magnitude and direction) a distance z above the midpoint between two equal charges +q+q that are a distance d apart (Figure 5.20). Check that your result is consistent with what you’d expect when zdzd.

(b) The same as part (a), only this time make the right-hand charge qq instead of +q+q.

Point P is a distance z above the midpoint between two charges separated by a horizontal distance d. The distance from each charge to point P is r, and the angle between r and the vertical is theta.
Figure 5.20 Finding the field of two identical source charges at the point P. Due to the symmetry, the net field at P is entirely vertical. (Notice that this is not true away from the midline between the charges.)

Strategy We add the two fields as vectors, per Equation 5.4. Notice that the system (and therefore the field) is symmetrical about the vertical axis; as a result, the horizontal components of the field vectors cancel. This simplifies the math. Also, we take care to express our final answer in terms of only quantities that are given in the original statement of the problem: q, z, d, and constants (π,ε0).(π,ε0).

Solution

  1. By symmetry, the horizontal (x)-components of EE cancel (Figure 5.21); Ex=14πε0qr2sinθ14πε0qr2sinθ=0Ex=14πε0qr2sinθ14πε0qr2sinθ=0.
    Point P is a distance z above the midpoint between two charges separated by a horizontal distance d. The distance from each charge to point P is r, and the angle between r and the vertical is theta. The x and y components of the electric field are shown as arrows whose tails are at point P. Four arrows are shown, as follows: E sub x r points to the left, E sub x l points to the right, E sub y l points up, and E sub E y r points up.
    Figure 5.21 Note that the horizontal components of the electric fields from the two charges cancel each other out, while the vertical components add together.

    The vertical (z)-component is given by
    Ez=14πε0qr2cosθ+14πε0qr2cosθ=14πε02qr2cosθ.Ez=14πε0qr2cosθ+14πε0qr2cosθ=14πε02qr2cosθ.

    Since none of the other components survive, this is the entire electric field, and it points in the k^k^ direction. Notice that this calculation uses the principle of superposition; we calculate the fields of the two charges independently and then add them together.
    What we want to do now is replace the quantities in this expression that we don’t know (such as r), or can’t easily measure (such as cosθ)cosθ) with quantities that we do know, or can measure. In this case, by geometry,
    r2=z2+(d2)2r2=z2+(d2)2

    and
    cosθ=zr=z[z2+(d2)2]1/2.cosθ=zr=z[z2+(d2)2]1/2.

    Thus, substituting,
    E(z)=14πε02q[z2+(d2)2]z[z2+(d2)2]1/2k^.E(z)=14πε02q[z2+(d2)2]z[z2+(d2)2]1/2k^.

    Simplifying, the desired answer is
    E(z)=14πε02qz[z2+(d2)2]3/2k^.E(z)=14πε02qz[z2+(d2)2]3/2k^.
    (5.5)
  2. If the source charges are equal and opposite, the vertical components cancel because Ez=14πε0qr2cosθ14πε0qr2cosθ=0Ez=14πε0qr2cosθ14πε0qr2cosθ=0
    and we get, for the horizontal component of EE,
    E(z)=14πε0qr2sinθi^14πε0qr2sinθi^=14πε02qr2sinθi^=14πε02q[z2+(d2)2](d2)[z2+(d2)2]1/2i^.E(z)=14πε0qr2sinθi^14πε0qr2sinθi^=14πε02qr2sinθi^=14πε02q[z2+(d2)2](d2)[z2+(d2)2]1/2i^.

    This becomes
    E(z)=14πε0qd[z2+(d2)2]3/2i^.E(z)=14πε0qd[z2+(d2)2]3/2i^.
    (5.6)

Significance It is a very common and very useful technique in physics to check whether your answer is reasonable by evaluating it at extreme cases. In this example, we should evaluate the field expressions for the cases d=0d=0, zdzd, and zz, and confirm that the resulting expressions match our physical expectations. Let’s do so:

Let’s start with Equation 5.5, the field of two identical charges. From far away (i.e., zd),zd), the two source charges should “merge” and we should then “see” the field of just one charge, of size 2q. So, let zd;zd; then we can neglect d2d2 in Equation 5.5 to obtain

limd0E=14πε02qz[z2]3/2k^=14πε02qzz3k^=14πε0(2q)z2k^,limd0E=14πε02qz[z2]3/2k^=14πε02qzz3k^=14πε0(2q)z2k^,

which is the correct expression for a field at a distance z away from a charge 2q.

Next, we consider the field of equal and opposite charges, Equation 5.6. It can be shown (via a Taylor expansion) that for dzdz, this becomes

E(z)=14πε0qdz3i^,E(z)=14πε0qdz3i^,
(5.7)

which is the field of a dipole, a system that we will study in more detail later. (Note that the units of EE are still correct in this expression, since the units of d in the numerator cancel the unit of the “extra” z in the denominator.) If z is very large (z)(z), then E0E0, as it should; the two charges “merge” and so cancel out.

Check Your Understanding 5.3

What is the electric field due to a single point particle?

Interactive

Try this simulation of electric field hockey to get the charge in the goal by placing other charges on the field.

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