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University Physics Volume 2

7.3 Calculations of Electric Potential

University Physics Volume 27.3 Calculations of Electric Potential
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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:
  • Calculate the potential due to a point charge
  • Calculate the potential of a system of multiple point charges
  • Describe an electric dipole
  • Define dipole moment
  • Calculate the potential of a continuous charge distribution

Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (such as charge on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider.

We can use calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge q. Noting the connection between work and potential W=qΔV,W=qΔV, as in the last section, we can obtain the following result.

Electric Potential V of a Point Charge

The electric potential V of a point charge is given by

V=kqr(point charge)V=kqr(point charge)
(7.8)

where k is a constant equal to 8.99×109N·m2/C2.8.99×109N·m2/C2.

The potential at infinity is chosen to be zero. Thus, V for a point charge decreases with distance, whereas EE for a point charge decreases with distance squared:

E=Fqt=kqr2.E=Fqt=kqr2.

Recall that the electric potential V is a scalar and has no direction, whereas the electric field EE is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that V is closely associated with energy, a scalar, whereas EE is closely associated with force, a vector.

Example 7.10

What Voltage Is Produced by a Small Charge on a Metal Sphere? Charges in static electricity are typically in the nanocoulomb (nC) to microcoulomb (μC)(μC) range. What is the voltage 5.00 cm away from the center of a 1-cm-diameter solid metal sphere that has a –3.00-nC static charge?

Strategy As we discussed in Electric Charges and Fields, charge on a metal sphere spreads out uniformly and produces a field like that of a point charge located at its center. Thus, we can find the voltage using the equation V=kqr.V=kqr.

Solution Entering known values into the expression for the potential of a point charge, we obtain

V=kqr=(8.99×109N·m2/C2)(−3.00×10−9C5.00×10−2m)=−539V.V=kqr=(8.99×109N·m2/C2)(−3.00×10−9C5.00×10−2m)=−539V.

Significance The negative value for voltage means a positive charge would be attracted from a larger distance, since the potential is lower (more negative) than at larger distances. Conversely, a negative charge would be repelled, as expected.

Example 7.11

What Is the Excess Charge on a Van de Graaff Generator? A demonstration Van de Graaff generator has a 25.0-cm-diameter metal sphere that produces a voltage of 100 kV near its surface (Figure 7.18). What excess charge resides on the sphere? (Assume that each numerical value here is shown with three significant figures.)

The figure shows the parts of Van de Graaff generator.
Figure 7.18 The voltage of this demonstration Van de Graaff generator is measured between the charged sphere and ground. Earth’s potential is taken to be zero as a reference. The potential of the charged conducting sphere is the same as that of an equal point charge at its center.

Strategy The potential on the surface is the same as that of a point charge at the center of the sphere, 12.5 cm away. (The radius of the sphere is 12.5 cm.) We can thus determine the excess charge using the equation

V=kqr.V=kqr.

Solution Solving for q and entering known values gives

q=rVk=(0.125m)(100×103V)8.99×109N·m2/C2=1.39×10−6C=1.39μC.q=rVk=(0.125m)(100×103V)8.99×109N·m2/C2=1.39×10−6C=1.39μC.

Significance This is a relatively small charge, but it produces a rather large voltage. We have another indication here that it is difficult to store isolated charges.

Check Your Understanding 7.8

What is the potential inside the metal sphere in Example 7.10?

The voltages in both of these examples could be measured with a meter that compares the measured potential with ground potential. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potential difference between two points that is of importance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. As noted earlier, this is analogous to taking sea level as h=0h=0 when considering gravitational potential energy Ug=mghUg=mgh.

Systems of Multiple Point Charges

Just as the electric field obeys a superposition principle, so does the electric potential. Consider a system consisting of N charges q1,q2,,qN.q1,q2,,qN. What is the net electric potential V at a space point P from these charges? Each of these charges is a source charge that produces its own electric potential at point P, independent of whatever other changes may be doing. Let V1,V2,,VNV1,V2,,VN be the electric potentials at P produced by the charges q1,q2,,qN,q1,q2,,qN, respectively. Then, the net electric potential VPVP at that point is equal to the sum of these individual electric potentials. You can easily show this by calculating the potential energy of a test charge when you bring the test charge from the reference point at infinity to point P:

VP=V1+V2++VN=1NVi.VP=V1+V2++VN=1NVi.

Note that electric potential follows the same principle of superposition as electric field and electric potential energy. To show this more explicitly, note that a test charge qiqi at the point P in space has distances of r1,r2,,rNr1,r2,,rN from the N charges fixed in space above, as shown in Figure 7.19. Using our formula for the potential of a point charge for each of these (assumed to be point) charges, we find that

VP=1Nkqiri=k1Nqiri.VP=1Nkqiri=k1Nqiri.
(7.9)

Therefore, the electric potential energy of the test charge is

UP=qtVP=qtk1Nqiri,UP=qtVP=qtk1Nqiri,

which is the same as the work to bring the test charge into the system, as found in the first section of the chapter.

The figure shows N charges located at different distances from a fixed point P.
Figure 7.19 Notation for direct distances from charges to a space point P.

The Electric Dipole

An electric dipole is a system of two equal but opposite charges a fixed distance apart. This system is used to model many real-world systems, including atomic and molecular interactions. One of these systems is the water molecule, under certain circumstances. These circumstances are met inside a microwave oven, where electric fields with alternating directions make the water molecules change orientation. This vibration is the same as heat at the molecular level.

Example 7.12

Electric Potential of a Dipole Consider the dipole in Figure 7.20 with the charge magnitude of q=3.0nCq=3.0nC and separation distance d=4.0cm.d=4.0cm. What is the potential at the following locations in space? (a) (0, 0, 1.0 cm); (b) (0, 0, –5.0 cm); (c) (3.0 cm, 0, 2.0 cm).

The figure shows an electric dipole with two charges (3.0nC and -3.0nC) located 4.0cm apart on the z axis. The center of the dipole is at the origin and three other points are marked at (0, 0, 1.0 cm), (0, 0, –5.0 cm) and (3.0 cm, 0, 2.0 cm).
Figure 7.20 A general diagram of an electric dipole, and the notation for the distances from the individual charges to a point P in space.

Strategy Apply VP=k1NqiriVP=k1Nqiri to each of these three points.

Solution

  1. VP=k1Nqiri=(9.0×109N·m2/C2)(3.0nC0.010m3.0nC0.030m)=1.8×103VVP=k1Nqiri=(9.0×109N·m2/C2)(3.0nC0.010m3.0nC0.030m)=1.8×103V
  2. VP=k1Nqiri=(9.0×109N·m2/C2)(3.0nC0.070m3.0nC0.030m)=−5.1×102VVP=k1Nqiri=(9.0×109N·m2/C2)(3.0nC0.070m3.0nC0.030m)=−5.1×102V
  3. VP=k1Nqiri=(9.0×109N·m2/C2)(3.0nC0.030m3.0nC0.050m)=3.6×102VVP=k1Nqiri=(9.0×109N·m2/C2)(3.0nC0.030m3.0nC0.050m)=3.6×102V

Significance Note that evaluating potential is significantly simpler than electric field, due to potential being a scalar instead of a vector.

Check Your Understanding 7.9

What is the potential on the x-axis? The z-axis?

Now let us consider the special case when the distance of the point P from the dipole is much greater than the distance between the charges in the dipole, rd;rd; for example, when we are interested in the electric potential due to a polarized molecule such as a water molecule. This is not so far (infinity) that we can simply treat the potential as zero, but the distance is great enough that we can simplify our calculations relative to the previous example.

We start by noting that in Figure 7.21 the potential is given by

VP=V++V=k(qr+qr)VP=V++V=k(qr+qr)

where

r±=x2+(zd2)2.r±=x2+(zd2)2.
The figure shows an electric dipole located on the z axis with center at the origin. Point P, located at (x, 0, z) is distance r away from the origin.
Figure 7.21 A general diagram of an electric dipole, and the notation for the distances from the individual charges to a point P in space.

This is still the exact formula. To take advantage of the fact that rd,rd, we rewrite the radii in terms of polar coordinates, with x=rsinθx=rsinθ and z=rcosθz=rcosθ. This gives us

r±=r2sin2θ+(rcosθd2)2.r±=r2sin2θ+(rcosθd2)2.

We can simplify this expression by pulling r out of the root,

r±=rsin2θ+(cosθd2r)2r±=rsin2θ+(cosθd2r)2

and then multiplying out the parentheses

r±=rsin2θ+cos2θcosθdr+(d2r)2=r1cosθdr+(d2r)2.r±=rsin2θ+cos2θcosθdr+(d2r)2=r1cosθdr+(d2r)2.

The last term in the root is small enough to be negligible (remember rd,rd, and hence (d/r)2(d/r)2 is extremely small, effectively zero to the level we will probably be measuring), leaving us with

r±=r1cosθdr.r±=r1cosθdr.

Using the binomial approximation (a standard result from the mathematics of series, when αα is small)

11α1±α211α1±α2

and substituting this into our formula for VPVP , we get

VP=k[qr(1+dcosθ2r)qr(1dcosθ2r)]=kqdcosθr2.VP=k[qr(1+dcosθ2r)qr(1dcosθ2r)]=kqdcosθr2.

This may be written more conveniently if we define a new quantity, the electric dipole moment,

p=qd,p=qd,
(7.10)

where these vectors point from the negative to the positive charge. Note that this has magnitude qd. This quantity allows us to write the potential at point P due to a dipole at the origin as

VP=kp·r^r2.VP=kp·r^r2.
(7.11)

A diagram of the application of this formula is shown in Figure 7.22.

The figure shows two vectors r and p with an angle theta between them.
Figure 7.22 The geometry for the application of the potential of a dipole.

There are also higher-order moments, for quadrupoles, octupoles, and so on. You will see these in future classes.

Potential of Continuous Charge Distributions

We have been working with point charges a great deal, but what about continuous charge distributions? Recall from Equation 7.9 that

VP=kqiri.VP=kqiri.

We may treat a continuous charge distribution as a collection of infinitesimally separated individual points. This yields the integral

VP=kdqrVP=kdqr
(7.12)

for the potential at a point P. Note that r is the distance from each individual point in the charge distribution to the point P. As we saw in Electric Charges and Fields, the infinitesimal charges are given by

dq={λdl(one dimension)σdA(two dimensions)ρdV(three dimensions)dq={λdl(one dimension)σdA(two dimensions)ρdV(three dimensions)

where λλ is linear charge density, σσ is the charge per unit area, and ρρ is the charge per unit volume.

Example 7.13

Potential of a Line of Charge Find the electric potential of a uniformly charged, nonconducting wire with linear density λλ (coulomb/meter) and length L at a point that lies on a line that divides the wire into two equal parts.

Strategy To set up the problem, we choose Cartesian coordinates in such a way as to exploit the symmetry in the problem as much as possible. We place the origin at the center of the wire and orient the y-axis along the wire so that the ends of the wire are at y=±L/2y=±L/2. The field point P is in the xy-plane and since the choice of axes is up to us, we choose the x-axis to pass through the field point P, as shown in Figure 7.23.

The figure shows a line charge on the y-axis with its center at the origin. Point P is located on the x-axis at distance x away from the origin.
Figure 7.23 We want to calculate the electric potential due to a line of charge.

Solution Consider a small element of the charge distribution between y and y+dyy+dy. The charge in this cell is dq=λdydq=λdy and the distance from the cell to the field point P is x2+y2.x2+y2. Therefore, the potential becomes

VP=kdqr=kL/2L/2λdyx2+y2=kλ[ln(y+y2+x2)]L/2L/2=kλ[ln((L2)+(L2)2+x2)ln((L2)+(L2)2+x2)]=kλln[L+L2+4x2L+L2+4x2].VP=kdqr=kL/2L/2λdyx2+y2=kλ[ln(y+y2+x2)]L/2L/2=kλ[ln((L2)+(L2)2+x2)ln((L2)+(L2)2+x2)]=kλln[L+L2+4x2L+L2+4x2].

Significance Note that this was simpler than the equivalent problem for electric field, due to the use of scalar quantities. Recall that we expect the zero level of the potential to be at infinity, when we have a finite charge. To examine this, we take the limit of the above potential as x approaches infinity; in this case, the terms inside the natural log approach one, and hence the potential approaches zero in this limit. Note that we could have done this problem equivalently in cylindrical coordinates; the only effect would be to substitute r for x and z for y.

Example 7.14

Potential Due to a Ring of Charge A ring has a uniform charge density λλ, with units of coulomb per unit meter of arc. Find the electric potential at a point on the axis passing through the center of the ring.

Strategy We use the same procedure as for the charged wire. The difference here is that the charge is distributed on a circle. We divide the circle into infinitesimal elements shaped as arcs on the circle and use cylindrical coordinates shown in Figure 7.24.

The figure shows a ring of charge located on the xy-plane with its center at the origin. Point P is located on the z-axis at distance z away from the origin.
Figure 7.24 We want to calculate the electric potential due to a ring of charge.

Solution A general element of the arc between θθ and θ+dθθ+dθ is of length RdθRdθ and therefore contains a charge equal to λRdθ.λRdθ. The element is at a distance of z2+R2z2+R2 from P, and therefore the potential is

VP=kdqr=k02πλRdθz2+R2=kλRz2+R202πdθ=2πkλRz2+R2=kqtotz2+R2.VP=kdqr=k02πλRdθz2+R2=kλRz2+R202πdθ=2πkλRz2+R2=kqtotz2+R2.

Significance This result is expected because every element of the ring is at the same distance from point P. The net potential at P is that of the total charge placed at the common distance, z2+R2z2+R2.

Example 7.15

Potential Due to a Uniform Disk of Charge A disk of radius R has a uniform charge density σσ, with units of coulomb meter squared. Find the electric potential at any point on the axis passing through the center of the disk.

Strategy We divide the disk into ring-shaped cells, and make use of the result for a ring worked out in the previous example, then integrate over r in addition to θθ. This is shown in Figure 7.25.

The figure shows a disk of charge located on the xy-plane with its center at the origin. Point P is located on the z-axis at distance z away from the origin.
Figure 7.25 We want to calculate the electric potential due to a disk of charge.

Solution An infinitesimal width cell between cylindrical coordinates r and r+drr+dr shown in Figure 7.25 will be a ring of charges whose electric potential dVPdVP at the field point has the following expression

dVP=kdqz2+r2dVP=kdqz2+r2

where

dq=σ2πrdr.dq=σ2πrdr.

The superposition of potential of all the infinitesimal rings that make up the disk gives the net potential at point P. This is accomplished by integrating from r=0r=0 to r=Rr=R:

VP=dVP=k2πσ0Rrdrz2+r2,=k2πσ(z2+R2z2).VP=dVP=k2πσ0Rrdrz2+r2,=k2πσ(z2+R2z2).

Significance The basic procedure for a disk is to first integrate around θθ and then over r. This has been demonstrated for uniform (constant) charge density. Often, the charge density will vary with r, and then the last integral will give different results.

Example 7.16

Potential Due to an Infinite Charged Wire Find the electric potential due to an infinitely long uniformly charged wire.

Strategy Since we have already worked out the potential of a finite wire of length L in Example 7.7, we might wonder if taking LL in our previous result will work:

VP=limLkλln(L+L2+4x2L+L2+4x2).VP=limLkλln(L+L2+4x2L+L2+4x2).

However, this limit does not exist because the argument of the logarithm becomes [2/0] as LL, so this way of finding V of an infinite wire does not work. The reason for this problem may be traced to the fact that the charges are not localized in some space but continue to infinity in the direction of the wire. Hence, our (unspoken) assumption that zero potential must be an infinite distance from the wire is no longer valid.

To avoid this difficulty in calculating limits, let us use the definition of potential by integrating over the electric field from the previous section, and the value of the electric field from this charge configuration from the previous chapter.

Solution We use the integral

VP=RPE·dlVP=RPE·dl

where R is a finite distance from the line of charge, as shown in Figure 7.26.

The figure shows an infinite line charge on the z-axis. Points P and R are located on the x-axis.
Figure 7.26 Points of interest for calculating the potential of an infinite line of charge.

With this setup, we use EP=2kλ1ss^EP=2kλ1ss^ and dl=dsdl=ds to obtain

VPVR=RP2kλ1sds=−2kλlnsPsR.VPVR=RP2kλ1sds=−2kλlnsPsR.

Now, if we define the reference potential VR=0VR=0 at sR=1m,sR=1m, this simplifies to

VP=−2kλlnsP.VP=−2kλlnsP.

Note that this form of the potential is quite usable; it is 0 at 1 m and is undefined at infinity, which is why we could not use the latter as a reference.

Significance Although calculating potential directly can be quite convenient, we just found a system for which this strategy does not work well. In such cases, going back to the definition of potential in terms of the electric field may offer a way forward.

Check Your Understanding 7.10

What is the potential on the axis of a nonuniform ring of charge, where the charge density is λ(θ)=λcosθλ(θ)=λcosθ ?

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