Skip to Content
OpenStax Logo
University Physics Volume 2

5.5 Calculating Electric Fields of Charge Distributions

University Physics Volume 25.5 Calculating Electric Fields of Charge Distributions
Buy book
  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 what a continuous source charge distribution is and how it is related to the concept of quantization of charge
  • Describe line charges, surface charges, and volume charges
  • Calculate the field of a continuous source charge distribution of either sign

The charge distributions we have seen so far have been discrete: made up of individual point particles. This is in contrast with a continuous charge distribution, which has at least one nonzero dimension. If a charge distribution is continuous rather than discrete, we can generalize the definition of the electric field. We simply divide the charge into infinitesimal pieces and treat each piece as a point charge.

Note that because charge is quantized, there is no such thing as a “truly” continuous charge distribution. However, in most practical cases, the total charge creating the field involves such a huge number of discrete charges that we can safely ignore the discrete nature of the charge and consider it to be continuous. This is exactly the kind of approximation we make when we deal with a bucket of water as a continuous fluid, rather than a collection of H2OH2O molecules.

Our first step is to define a charge density for a charge distribution along a line, across a surface, or within a volume, as shown in Figure 5.22.

Figure a shows a long rod with linear charge density lambda. A small segment of the rod is shaded and labeled d l. Figure b shows a surface with surface charge density sigma. A small area within the surface is shaded and labeled d A. Figure c shows a volume with volume charge density rho. A small volume within it is shaded and labeled d V. Figure d shows a surface with two regions shaded and labeled q 1 and q2. A point P is identified above (not on) the surface. A thin line indicates the distance from each of the shaded regions. The vectors E 1 and E 2 are drawn at point P and point away from the respective shaded region. E net is the vector sum of E 1 and E 2. In this case, it points up, away from the surface.
Figure 5.22 The configuration of charge differential elements for a (a) line charge, (b) sheet of charge, and (c) a volume of charge. Also note that (d) some of the components of the total electric field cancel out, with the remainder resulting in a net electric field.

Definitions of charge density:

  • λλ charge per unit length (linear charge density); units are coulombs per meter (C/m)
  • σσ charge per unit area (surface charge density); units are coulombs per square meter (C/m2)(C/m2)
  • ρρ charge per unit volume (volume charge density); units are coulombs per cubic meter (C/m3)(C/m3)

Then, for a line charge, a surface charge, and a volume charge, the summation in Equation 5.4 becomes an integral and qiqi is replaced by dq=λdldq=λdl, σdAσdA, or ρdVρdV, respectively:

Point charges:E(P)=14πε0i=1N(qir2)r^Point charges:E(P)=14πε0i=1N(qir2)r^
(5.8)
Line charge:E(P)=14πε0line(λdlr2)r^Line charge:E(P)=14πε0line(λdlr2)r^
(5.9)
Surface charge:E(P)=14πε0surface(σdAr2)r^Surface charge:E(P)=14πε0surface(σdAr2)r^
(5.10)
Volume charge:E(P)=14πε0volume(ρdVr2)r^Volume charge:E(P)=14πε0volume(ρdVr2)r^
(5.11)

The integrals are generalizations of the expression for the field of a point charge. They implicitly include and assume the principle of superposition. The “trick” to using them is almost always in coming up with correct expressions for dl, dA, or dV, as the case may be, expressed in terms of r, and also expressing the charge density function appropriately. It may be constant; it might be dependent on location.

Note carefully the meaning of r in these equations: It is the distance from the charge element (qi,λdl,σdA,ρdV)(qi,λdl,σdA,ρdV) to the location of interest, P(x,y,z)P(x,y,z) (the point in space where you want to determine the field). However, don’t confuse this with the meaning of r^r^; we are using it and the vector notation EE to write three integrals at once. That is, Equation 5.9 is actually

Ex(P)=14πε0line(λdlr2)x,Ey(P)=14πε0line(λdlr2)y,Ez(P)=14πε0line(λdlr2)z.Ex(P)=14πε0line(λdlr2)x,Ey(P)=14πε0line(λdlr2)y,Ez(P)=14πε0line(λdlr2)z.

Example 5.5

Electric Field of a Line Segment Find the electric field a distance z above the midpoint of a straight line segment of length L that carries a uniform line charge density λλ.

Strategy Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces of length dl, each of which carries a differential amount of charge dq=λdldq=λdl. Then, we calculate the differential field created by two symmetrically placed pieces of the wire, using the symmetry of the setup to simplify the calculation (Figure 5.23). Finally, we integrate this differential field expression over the length of the wire (half of it, actually, as we explain below) to obtain the complete electric field expression.

A long, thin wire is on the x axis. The end of the wire is a distance z from the center of the wire. A small segment of the wire, a distance x to the right of the center of the wire, is shaded. Another segment, the same distance to the left of center, is also shaded. Point P is a distance z above the center of the wire, on the z axis. Point P is a distance r from each shaded region. The r vectors point from each shaded region to point P. Vectors d E 1 and d E 2 are drawn at point P. d E 1 points away from the left side shaded region and points up and right, at an angle theta to the z axis. d E 2 points away from the right side shaded region and points up and r left, making the same angle with the vertical as d E 1. The two d E vectors are equal in length.
Figure 5.23 A uniformly charged segment of wire. The electric field at point P can be found by applying the superposition principle to symmetrically placed charge elements and integrating.

Solution Before we jump into it, what do we expect the field to “look like” from far away? Since it is a finite line segment, from far away, it should look like a point charge. We will check the expression we get to see if it meets this expectation.

The electric field for a line charge is given by the general expression

E(P)=14πε0lineλdlr2r^.E(P)=14πε0lineλdlr2r^.

The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal (x)-components of the field cancel, so that the net field points in the z-direction. Let’s check this formally.

The total field E(P)E(P) is the vector sum of the fields from each of the two charge elements (call them E1E1 and E2E2, for now):

E(P)=E1+E2=E1xi^+E1zk^+E2x(i^)+E2zk^.E(P)=E1+E2=E1xi^+E1zk^+E2x(i^)+E2zk^.

Because the two charge elements are identical and are the same distance away from the point P where we want to calculate the field, E1x=E2x,E1x=E2x, so those components cancel. This leaves

E(P)=E1zk^+E2zk^=E1cosθk^+E2cosθk^.E(P)=E1zk^+E2zk^=E1cosθk^+E2cosθk^.

These components are also equal, so we have

E(P)=14πε0λdlr2cosθk^+14πε0λdlr2cosθk^=14πε00L/22λdxr2cosθk^E(P)=14πε0λdlr2cosθk^+14πε0λdlr2cosθk^=14πε00L/22λdxr2cosθk^

where our differential line element dl is dx, in this example, since we are integrating along a line of charge that lies on the x-axis. (The limits of integration are 0 to L2L2, not L2L2 to +L2+L2, because we have constructed the net field from two differential pieces of charge dq. If we integrated along the entire length, we would pick up an erroneous factor of 2.)

In principle, this is complete. However, to actually calculate this integral, we need to eliminate all the variables that are not given. In this case, both r and θθ change as we integrate outward to the end of the line charge, so those are the variables to get rid of. We can do that the same way we did for the two point charges: by noticing that

r=(z2+x2)1/2r=(z2+x2)1/2

and

cosθ=zr=z(z2+x2)1/2.cosθ=zr=z(z2+x2)1/2.

Substituting, we obtain

E(P)=14πε00L/22λdx(z2+x2)z(z2+x2)1/2k^ =14πε00L/22λz(z2+x2)3/2dxk^ =2λz4πε0 [ x z2z2+x2 ] | 0 L/2 k^ E(P)=14πε00L/22λdx(z2+x2)z(z2+x2)1/2k^ =14πε00L/22λz(z2+x2)3/2dxk^ =2λz4πε0 [ x z2z2+x2 ] | 0 L/2 k^

which simplifies to

E(z)=14πε0λLzz2+L24k^.E(z)=14πε0λLzz2+L24k^.
(5.12)

Significance Notice, once again, the use of symmetry to simplify the problem. This is a very common strategy for calculating electric fields. The fields of nonsymmetrical charge distributions have to be handled with multiple integrals and may need to be calculated numerically by a computer.

Check Your Understanding 5.4

How would the strategy used above change to calculate the electric field at a point a distance z above one end of the finite line segment?

Example 5.6

Electric Field of an Infinite Line of Charge Find the electric field a distance z above the midpoint of an infinite line of charge that carries a uniform line charge density λλ.

Strategy This is exactly like the preceding example, except the limits of integration will be to ++.

Solution Again, the horizontal components cancel out, so we wind up with

E(P)=14πε0λdxr2cosθk^E(P)=14πε0λdxr2cosθk^

where our differential line element dl is dx, in this example, since we are integrating along a line of charge that lies on the x-axis. Again,

cosθ=zr=z(z2+x2)1/2.cosθ=zr=z(z2+x2)1/2.

Substituting, we obtain

E(P)=14πε0λdx(z2+x2)z(z2+x2)1/2k^=14πε0λz(z2+x2)3/2dxk^=λz4πε0[xz2z2+x2]|k^,E(P)=14πε0λdx(z2+x2)z(z2+x2)1/2k^=14πε0λz(z2+x2)3/2dxk^=λz4πε0[xz2z2+x2]|k^,

which simplifies to

E(z)=14πε02λzk^.E(z)=14πε02λzk^.

Significance Our strategy for working with continuous charge distributions also gives useful results for charges with infinite dimension.

In the case of a finite line of charge, note that for zLzL, z2z2 dominates the L in the denominator, so that Equation 5.12 simplifies to

E14πε0λLz2k^.E14πε0λLz2k^.

If you recall that λL=qλL=q, the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected.

In the limit LL, on the other hand, we get the field of an infinite straight wire, which is a straight wire whose length is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the field is to be calculated:

E(z)=14πε02λzk^.E(z)=14πε02λzk^.
(5.13)

An interesting artifact of this infinite limit is that we have lost the usual 1/r21/r2 dependence that we are used to. This will become even more intriguing in the case of an infinite plane.

Example 5.7

Electric Field 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 polar coordinates shown in Figure 5.24.

A ring of radius R is shown in the x y plane of an x y z coordinate system. The ring is centered on the origin. A small segment of the ring is shaded. The segment is at an angle of theta from the x axis, subtends an angle of d theta, and contains a charge of d q equal to lambda R d theta. Point P is on the z axis, a distance of z above the center of the ring. The distance from the shaded segment to point P is equal to the square root of R squared plus squared.
Figure 5.24 The system and variable for calculating the electric field due to a ring of charge.

Solution The electric field for a line charge is given by the general expression

E(P)=14πε0lineλdlr2r^.E(P)=14πε0lineλdlr2r^.

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 r=z2+R2r=z2+R2 from P, the angle is cosϕ=zz2+R2cosϕ=zz2+R2, and therefore the electric field is

E(P)=14πε0lineλdlr2r^=14πε002πλRdθz2+R2zz2+R2z^=14πε0λRz(z2+R2)3/2z^02πdθ=14πε02πλRz(z2+R2)3/2z^=14πε0qtotz(z2+R2)3/2z^.E(P)=14πε0lineλdlr2r^=14πε002πλRdθz2+R2zz2+R2z^=14πε0λRz(z2+R2)3/2z^02πdθ=14πε02πλRz(z2+R2)3/2z^=14πε0qtotz(z2+R2)3/2z^.

Significance As usual, symmetry simplified this problem, in this particular case resulting in a trivial integral. Also, when we take the limit of z>>Rz>>R, we find that

E14πε0qtotz2z^,E14πε0qtotz2z^,

as we expect.

Example 5.8

The Field of a Disk Find the electric field of a circular thin disk of radius R and uniform charge density at a distance z above the center of the disk (Figure 5.25)

A disk of radius R is shown in the x y plane of an x y z coordinate system. The disk is centered on the origin. A ring, concentric with the disk, of radius r prime and width d r prime is indicated and two small segments on opposite sides of the ring are shaded and labeled as having charge d q. The test point is on the z axis, a distance of z above the center of the disk. The distance from each shaded segment to the test point is r. The electric field contributions, d E, due to the d q charges are shown as arrows in the directions of the associated r vectors. The d E vectors are at an angle of theta to the z axis.
Figure 5.25 A uniformly charged disk. As in the line charge example, the field above the center of this disk can be calculated by taking advantage of the symmetry of the charge distribution.

Strategy The electric field for a surface charge is given by

E(P)=14πε0surfaceσdAr2r^.E(P)=14πε0surfaceσdAr2r^.

To solve surface charge problems, we break the surface into symmetrical differential “stripes” that match the shape of the surface; here, we’ll use rings, as shown in the figure. Again, by symmetry, the horizontal components cancel and the field is entirely in the vertical (k^)(k^) direction. The vertical component of the electric field is extracted by multiplying by cosθcosθ, so

E(P)=14πε0surfaceσdAr2cosθk^.E(P)=14πε0surfaceσdAr2cosθk^.

As before, we need to rewrite the unknown factors in the integrand in terms of the given quantities. In this case,

dA=2πrdrr2=r2+z2cosθ=z(r2+z2)1/2.dA=2πrdrr2=r2+z2cosθ=z(r2+z2)1/2.

(Please take note of the two different “r’s” here; r is the distance from the differential ring of charge to the point P where we wish to determine the field, whereas rr is the distance from the center of the disk to the differential ring of charge.) Also, we already performed the polar angle integral in writing down dA.

Solution Substituting all this in, we get

E(P)=E(z)=14πε00Rσ(2πrdr)z(r2+z2)3/2k^=14πε0(2πσz)(1z1R2+z2)k^E(P)=E(z)=14πε00Rσ(2πrdr)z(r2+z2)3/2k^=14πε0(2πσz)(1z1R2+z2)k^

or, more simply,

E(z)=14πε0(2πσ2πσzR2+z2)k^.E(z)=14πε0(2πσ2πσzR2+z2)k^.
(5.14)

Significance Again, it can be shown (via a Taylor expansion) that when zRzR, this reduces to

E(z)14πε0σπR2z2k^,E(z)14πε0σπR2z2k^,

which is the expression for a point charge Q=σπR2.Q=σπR2.

Check Your Understanding 5.5

How would the above limit change with a uniformly charged rectangle instead of a disk?

As RR, Equation 5.14 reduces to the field of an infinite plane, which is a flat sheet whose area is much, much greater than its thickness, and also much, much greater than the distance at which the field is to be calculated:

E=σ2ε0k^.E=σ2ε0k^.
(5.15)

Note that this field is constant. This surprising result is, again, an artifact of our limit, although one that we will make use of repeatedly in the future. To understand why this happens, imagine being placed above an infinite plane of constant charge. Does the plane look any different if you vary your altitude? No—you still see the plane going off to infinity, no matter how far you are from it. It is important to note that Equation 5.15 is because we are above the plane. If we were below, the field would point in the k^k^ direction.

Example 5.9

The Field of Two Infinite Planes Find the electric field everywhere resulting from two infinite planes with equal but opposite charge densities (Figure 5.26).

The figure shows two vertically oriented parallel plates A and B separated by a distance d. Plate A is positively charged and B is negatively charged. Electric field lines are parallel between the plates and curved outward 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.
Figure 5.26 Two charged infinite planes. Note the direction of the electric field.

Strategy We already know the electric field resulting from a single infinite plane, so we may use the principle of superposition to find the field from two.

Solution The electric field points away from the positively charged plane and toward the negatively charged plane. Since the σσ are equal and opposite, this means that in the region outside of the two planes, the electric fields cancel each other out to zero.

However, in the region between the planes, the electric fields add, and we get

E=σε0i^E=σε0i^

for the electric field. The i^i^ is because in the figure, the field is pointing in the +x-direction.

Significance Systems that may be approximated as two infinite planes of this sort provide a useful means of creating uniform electric fields.

Check Your Understanding 5.6

What would the electric field look like in a system with two parallel positively charged planes with equal charge densities?

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction
Citation information

© Oct 6, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.