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University Physics Volume 2

7.1 Electric Potential Energy

University Physics Volume 27.1 Electric Potential Energy
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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:
  • Define the work done by an electric force
  • Define electric potential energy
  • Apply work and potential energy in systems with electric charges

When a free positive charge q is accelerated by an electric field, it is given kinetic energy (Figure 7.2). The process is analogous to an object being accelerated by a gravitational field, as if the charge were going down an electrical hill where its electric potential energy is converted into kinetic energy, although of course the sources of the forces are very different. Let us explore the work done on a charge q by the electric field in this process, so that we may develop a definition of electric potential energy.

The first part of the figure shows two charged plates – one positive and one negative. A positive charge q is located between the plates and moves from point A to B. The second part of the figure shows a mass m rolling down a hill.
Figure 7.2 A charge accelerated by an electric field is analogous to a mass going down a hill. In both cases, potential energy decreases as kinetic energy increases, ΔU=ΔKΔU=ΔK. Work is done by a force, but since this force is conservative, we can write W=ΔUW=ΔU.

The electrostatic or Coulomb force is conservative, which means that the work done on q is independent of the path taken, as we will demonstrate later. This is exactly analogous to the gravitational force. When a force is conservative, it is possible to define a potential energy associated with the force. It is usually easier to work with the potential energy (because it depends only on position) than to calculate the work directly.

To show this explicitly, consider an electric charge +q+q fixed at the origin and move another charge +Q+Q toward q in such a manner that, at each instant, the applied force FF exactly balances the electric force FeFe on Q (Figure 7.3). The work done by the applied force FF on the charge Q changes the potential energy of Q. We call this potential energy the electrical potential energy of Q.

The figure shows two positive charges – fixed charge q and moving test charge Q and the forces on Q when is moved closer to q, from point P subscript 1 to point P subscript 2.
Figure 7.3 Displacement of “test” charge Q in the presence of fixed “source” charge q.

The work W12W12 done by the applied force FF when the particle moves from P1P1 to P2P2 may be calculated by

W12=P1P2F·dl.W12=P1P2F·dl.

Since the applied force FF balances the electric force FeFe on Q, the two forces have equal magnitude and opposite directions. Therefore, the applied force is

F=Fe=kqQr2r^,F=Fe=kqQr2r^,

where we have defined positive to be pointing away from the origin and r is the distance from the origin. The directions of both the displacement and the applied force in the system in Figure 7.3 are parallel, and thus the work done on the system is positive.

We use the letter U to denote electric potential energy, which has units of joules (J). When a conservative force does negative work, the system gains potential energy. When a conservative force does positive work, the system loses potential energy, ΔU=W.ΔU=W. In the system in Figure 7.3, the Coulomb force acts in the opposite direction to the displacement; therefore, the work is negative. However, we have increased the potential energy in the two-charge system.

Example 7.1

Kinetic Energy of a Charged Particle A +3.0-nC+3.0-nC charge Q is initially at rest a distance of 10 cm (r1r1) from a +5.0-nC+5.0-nC charge q fixed at the origin (Figure 7.4). Naturally, the Coulomb force accelerates Q away from q, eventually reaching 15 cm (r2r2).

The figure shows two positive charges, q (+5.0nC) and Q (+3.0nC) and the repelling force on Q, marked as F subscript e. Q is located at r subscript 1 = 10cm and F subscript e vector is towards r subscript 2 = 15cm.
Figure 7.4 The charge Q is repelled by q, thus having work done on it and gaining kinetic energy.
  1. What is the work done by the electric field between r1r1 and r2r2?
  2. How much kinetic energy does Q have at r2r2?

Strategy Calculate the work with the usual definition. Since Q started from rest, this is the same as the kinetic energy.

Solution Integrating force over distance, we obtain

W12=r1r2F·dr=r1r2kqQr2dr=[kqQr]r1r2=kqQ[−1r2+1r1]=(8.99×109Nm2/C2)(5.0×10−9C)(3.0×10−9C)[−10.15m+10.10m]=4.5×10−7J.W12=r1r2F·dr=r1r2kqQr2dr=[kqQr]r1r2=kqQ[−1r2+1r1]=(8.99×109Nm2/C2)(5.0×10−9C)(3.0×10−9C)[−10.15m+10.10m]=4.5×10−7J.

This is also the value of the kinetic energy at r2.r2.

Significance Charge Q was initially at rest; the electric field of q did work on Q, so now Q has kinetic energy equal to the work done by the electric field.

Check Your Understanding 7.1

If Q has a mass of 4.00μg,4.00μg, what is the speed of Q at r2?r2?

In this example, the work W done to accelerate a positive charge from rest is positive and results from a loss in U, or a negative ΔUΔU. A value for U can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

Electric Potential Energy

Work W done to accelerate a positive charge from rest is positive and results from a loss in U, or a negative ΔUΔU. Mathematically,

W=ΔU.W=ΔU.
(7.1)

Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done by a conservative force and gives added insight regarding energy and energy transformation without the necessity of dealing with the force directly. It is much more common, for example, to use the concept of electric potential energy than to deal with the Coulomb force directly in real-world applications.

In polar coordinates with q at the origin and Q located at r, the displacement element vector is dl=r^drdl=r^dr and thus the work becomes

W12=kqQr1r21r2r^·r^dr=kqQ1r2kqQ1r1.W12=kqQr1r21r2r^·r^dr=kqQ1r2kqQ1r1.

Notice that this result only depends on the endpoints and is otherwise independent of the path taken. To explore this further, compare path P1P1 to P2P2 with path P1P3P4P2P1P3P4P2 in Figure 7.5.

The figure shows two positive charges, q and Q and the repelling force on Q. There are four points P subscript 1, P subscript 2, P subscript 3 and P subscript 4 where P subscript 1 P subscript 3 and P subscript 2 P subscript 4 form two concentric segments centered at q. The force on Q is perpendicular to direction of displacement when Q moves from P subscript 1 to P subscript 3 or P subscript 3 to P subscript 2.
Figure 7.5 Two paths for displacement P1P1 to P2.P2. The work on segments P1P3P1P3 and P4P2P4P2 are zero due to the electrical force being perpendicular to the displacement along these paths. Therefore, work on paths P1P2P1P2 and P1P3P4P2P1P3P4P2 are equal.

The segments P1P3P1P3 and P4P2P4P2 are arcs of circles centered at q. Since the force on Q points either toward or away from q, no work is done by a force balancing the electric force, because it is perpendicular to the displacement along these arcs. Therefore, the only work done is along segment P3P4,P3P4, which is identical to P1P2.P1P2.

One implication of this work calculation is that if we were to go around the path P1P3P4P2P1,P1P3P4P2P1, the net work would be zero (Figure 7.6). Recall that this is how we determine whether a force is conservative or not. Hence, because the electric force is related to the electric field by F=qEF=qE, the electric field is itself conservative. That is,

E·dl=0.E·dl=0.

Note that Q is a constant.

The figure shows two positive charges, q and Q and the repelling force on Q. There are four points P subscript 1, P subscript 2, P subscript 3 and P subscript 4 where P subscript 1 P subscript 3 and P subscript 2 P subscript 4 form two concentric segments centered at q. Q’s direction of displacement is P subscript 1 P subscript 3 P subscript 4 P subscript  2 P subscript 1.
Figure 7.6 A closed path in an electric field. The net work around this path is zero.

Another implication is that we may define an electric potential energy. Recall that the work done by a conservative force is also expressed as the difference in the potential energy corresponding to that force. Therefore, the work WrefWref to bring a charge from a reference point to a point of interest may be written as

Wref=rrefrF·dlWref=rrefrF·dl

and, by Equation 7.1, the difference in potential energy (U2U1)(U2U1) of the test charge Q between the two points is

ΔU=rrefrF·dl.ΔU=rrefrF·dl.

Therefore, we can write a general expression for the potential energy of two point charges (in spherical coordinates):

ΔU=rrefrkqQr2dr=[kqQr]rrefr=kqQ[1r1rref].ΔU=rrefrkqQr2dr=[kqQr]rrefr=kqQ[1r1rref].

We may take the second term to be an arbitrary constant reference level, which serves as the zero reference:

U(r)=kqQrUref.U(r)=kqQrUref.

A convenient choice of reference that relies on our common sense is that when the two charges are infinitely far apart, there is no interaction between them. (Recall the discussion of reference potential energy in Potential Energy and Conservation of Energy.) Taking the potential energy of this state to be zero removes the term UrefUref from the equation (just like when we say the ground is zero potential energy in a gravitational potential energy problem), and the potential energy of Q when it is separated from q by a distance r assumes the form

U(r)=kqQr(zero reference atr=).U(r)=kqQr(zero reference atr=).
(7.2)

This formula is symmetrical with respect to q and Q, so it is best described as the potential energy of the two-charge system.

Example 7.2

Potential Energy of a Charged Particle A +3.0-nC+3.0-nC charge Q is initially at rest a distance of 10 cm (r1r1) from a +5.0-nC+5.0-nC charge q fixed at the origin (Figure 7.7). Naturally, the Coulomb force accelerates Q away from q, eventually reaching 15 cm (r2r2).

The figure shows two positive charges, q (+5.0nC) and Q (+3.0nC) and the repelling force on Q, marked as F subscript e. Q is located at r subscript 1 = 10cm and F subscript e vector is towards r subscript 2 = 15cm.
Figure 7.7 The charge Q is repelled by q, thus having work done on it and losing potential energy.

What is the change in the potential energy of the two-charge system from r1r1 to r2?r2?

Strategy Calculate the potential energy with the definition given above: ΔU12=r1r2F·dr.ΔU12=r1r2F·dr. Since Q started from rest, this is the same as the kinetic energy.

Solution We have

ΔU12=r1r2F·dr=r1r2kqQr2dr=[kqQr]r1r2=kqQ[1r21r1]=(8.99×109Nm2/C2)(5.0×10−9C)(3.0×10−9C)[10.15m10.10m]=−4.5×10−7J.ΔU12=r1r2F·dr=r1r2kqQr2dr=[kqQr]r1r2=kqQ[1r21r1]=(8.99×109Nm2/C2)(5.0×10−9C)(3.0×10−9C)[10.15m10.10m]=−4.5×10−7J.

Significance The change in the potential energy is negative, as expected, and equal in magnitude to the change in kinetic energy in this system. Recall from Example 7.1 that the change in kinetic energy was positive.

Check Your Understanding 7.2

What is the potential energy of Q relative to the zero reference at infinity at r2r2 in the above example?

Due to Coulomb’s law, the forces due to multiple charges on a test charge Q superimpose; they may be calculated individually and then added. This implies that the work integrals and hence the resulting potential energies exhibit the same behavior. To demonstrate this, we consider an example of assembling a system of four charges.

Example 7.3

Assembling Four Positive Charges Find the amount of work an external agent must do in assembling four charges +2.0μC,+2.0μC,+3.0μC,+4.0μC,+3.0μC,+4.0μC, and +5.0μC+5.0μC at the vertices of a square of side 1.0 cm, starting each charge from infinity (Figure 7.8).

The figure shows a square with side length 1.0cm and four charges (2.0µC, 3.0µC, 4.0µC and 5.0µC) located at four corners.
Figure 7.8 How much work is needed to assemble this charge configuration?

Strategy We bring in the charges one at a time, giving them starting locations at infinity and calculating the work to bring them in from infinity to their final location. We do this in order of increasing charge.

Solution Step 1. First bring the +2.0-μC+2.0-μC charge to the origin. Since there are no other charges at a finite distance from this charge yet, no work is done in bringing it from infinity,

W1=0.W1=0.

Step 2. While keeping the +2.0-μC+2.0-μC charge fixed at the origin, bring the +3.0-μC+3.0-μC charge to (x,y,z)=(1.0cm,0,0)(x,y,z)=(1.0cm,0,0) (Figure 7.9). Now, the applied force must do work against the force exerted by the +2.0-μC+2.0-μC charge fixed at the origin. The work done equals the change in the potential energy of the +3.0-μC+3.0-μC charge:

W2=kq1q2r12=(9.0×109N·m2C2)(2.0×10−6C)(3.0×10−6C)1.0×10−2m=5.4J.W2=kq1q2r12=(9.0×109N·m2C2)(2.0×10−6C)(3.0×10−6C)1.0×10−2m=5.4J.
The figure shows a square with side length 1.0cm and two charges (2.0µC and 3.0µC) on adjacent corners.
Figure 7.9 Step 2. Work W2W2 to bring the +3.0-μC+3.0-μC charge from infinity.

Step 3. While keeping the charges of +2.0μC+2.0μC and +3.0μC+3.0μC fixed in their places, bring in the +4.0-μC+4.0-μC charge to (x,y,z)=(1.0cm,1.0cm,0)(x,y,z)=(1.0cm,1.0cm,0) (Figure 7.10). The work done in this step is

W3=kq1q3r13+kq2q3r23=(9.0×109N·m2C2)[(2.0×10−6C)(4.0×10−6C)2×10−2m+(3.0×10−6C)(4.0×10−6C)1.0×10−2m]=15.9J.W3=kq1q3r13+kq2q3r23=(9.0×109N·m2C2)[(2.0×10−6C)(4.0×10−6C)2×10−2m+(3.0×10−6C)(4.0×10−6C)1.0×10−2m]=15.9J.
The figure shows a square with side length 1.0cm and three charges (2.0µC, 3.0µC and 4.0µC) on three corners.
Figure 7.10 Step 3. The work W3W3 to bring the +4.0-μC+4.0-μC charge from infinity.

Step 4. Finally, while keeping the first three charges in their places, bring the +5.0-μC+5.0-μC charge to (x,y,z)=(0,1.0cm,0)(x,y,z)=(0,1.0cm,0) (Figure 7.11). The work done here is

W4=kq4[q1r14+q2r24+q3r34],=(9.0×109N·m2C2)(5.0×10−6C)[(2.0×10−6C)1.0×10−2m+(3.0×10−6C)2×10−2m+(4.0×10−6C)1.0×10−2m]=36.5J.W4=kq4[q1r14+q2r24+q3r34],=(9.0×109N·m2C2)(5.0×10−6C)[(2.0×10−6C)1.0×10−2m+(3.0×10−6C)2×10−2m+(4.0×10−6C)1.0×10−2m]=36.5J.
The figure shows a square with side length 1.0cm and four charges (2.0µC, 3.0µC, 4.0µC and 5.0µC) located at four corners.
Figure 7.11 Step 4. The work W4W4 to bring the +5.0-μC+5.0-μC charge from infinity.

Hence, the total work done by the applied force in assembling the four charges is equal to the sum of the work in bringing each charge from infinity to its final position:

WT=W1+W2+W3+W4=0+5.4J+15.9J+36.5J=57.8J.WT=W1+W2+W3+W4=0+5.4J+15.9J+36.5J=57.8J.

Significance The work on each charge depends only on its pairwise interactions with the other charges. No more complicated interactions need to be considered; the work on the third charge only depends on its interaction with the first and second charges, the interaction between the first and second charge does not affect the third.

Check Your Understanding 7.3

Is the electrical potential energy of two point charges positive or negative if the charges are of the same sign? Opposite signs? How does this relate to the work necessary to bring the charges into proximity from infinity?

Note that the electrical potential energy is positive if the two charges are of the same type, either positive or negative, and negative if the two charges are of opposite types. This makes sense if you think of the change in the potential energy ΔUΔU as you bring the two charges closer or move them farther apart. Depending on the relative types of charges, you may have to work on the system or the system would do work on you, that is, your work is either positive or negative. If you have to do positive work on the system (actually push the charges closer), then the energy of the system should increase. If you bring two positive charges or two negative charges closer, you have to do positive work on the system, which raises their potential energy. Since potential energy is proportional to 1/r, the potential energy goes up when r goes down between two positive or two negative charges.

On the other hand, if you bring a positive and a negative charge nearer, you have to do negative work on the system (the charges are pulling you), which means that you take energy away from the system. This reduces the potential energy. Since potential energy is negative in the case of a positive and a negative charge pair, the increase in 1/r makes the potential energy more negative, which is the same as a reduction in potential energy.

The result from Example 7.1 may be extended to systems with any arbitrary number of charges. In this case, it is most convenient to write the formula as

W12N=k2iNjNqiqjrijforij.W12N=k2iNjNqiqjrijforij.
(7.3)

The factor of 1/2 accounts for adding each pair of charges twice.

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