# Electric Fields Due to Continuous Charge Distributions

BEF 22903

Lecture 4

Electric Fields Due to Continuous Charge Distributions

LEARNING OUTCOMES

After completing this lesson you will be able to:

1. use Coulomb’s Law to calculate the electric field due to high-symmetry charge distributions such as a line charge, a ring of charge, a charged disc, and a pair of parallel plates.

Review of Coulomb’s Law

Before we begin our derivation of electric field expressions due to continuous charge distributions, let us briefly review some of the relationships developed in earlier lessons.

Recall that around every charged object there is an electric field, E, with field lines that originate on positive charges and terminate on negative charges. When another chargedobject of charge Q is brought into a field it will experience a force of attraction or repulsion according to the formula

F= Q1E

Figure 1. (a) Electric field at a distance r due to a point charge Q. (b) Force acting on a test charge Q1 due to electric field of charge Q.

When we speak of the force between two point charges, Q and Q1(say), we can calculate its magnitude withCoulomb's Law

The constant, k, in Coulomb's Law has the value

where

is called the permittivity of free space and represents the"willingness" of a regionto establish an electric field. This value changes with the presence of dielectrics. We can now express Coulomb's Law as

Substituting this expression for F into ourequation F =Q1E we getan equation that we can use to calculate the magnitude of theelectricfield around a point charge.

General Principle for Finding the Electric Field Due to aContinuous Charge Distribution

In the previous section we saw how a simple manipulation of Coulomb's Law led us to find the electric field due to a single point charge. What do we do if we have a continuous charge distribution?

We can find the electric field due to a continuous charge distribution by the method of superposition principle. According to this principle we can obtain the total electric field due to an object of finite size by summing up the electric field caused by each tiny, infinitesimal part of the charge distribution.

Consider the oddly-shaped charge object shown in Figure 2. To apply the superposition principle, we first divide the charge distribution into small elements, each of which containsdqand then calculate the electric field due to one of these elements at point P. The total field is obtained by summing the contributions of all the charge elements.

For the individual charge elements

Because the charge distribution is continuous
/
Figure 2

Concept of Charge Density

When we deal with continuous charge distributions, it is most convenient to express the charge on an object as a charge density rather than as a total charge.There are three different ways in which charge can be distributed:

1. Charge that is spread throughout a three-dimensional volume gives rise to a volume charge density, denoted bythe symbol “ρv” (just as we do for mass density).

[Eq. 1]

Since charge is measured in Coulombs [C], and volume is in meters3 [m3], the units of the electric charge density of Equation [1] are [C/m3]. Note that since electric charge can be negative or positive, the charge density can be negative, positive or zero.

Worked Example 1

Suppose we have +5 [C] of electric charge uniformly (evenly) spread within a sphere of radius 2 meters. What is the electric volume charge density in the sphere?

Solution

1. If an amount of charge is spread over some two-dimensional surface (which may be flat or curved), we have asurface charge density, and will use the symbol “” to represent such a distribution.

[Eq.2]

Since charge is measured in Coulombs [C], and area is in meters2 [m2], the units of the electric charge density of Equation [2] are [C/m2]. Similarly, like for volume charge density, since electric charge can be negative or positive, the surface charge density can be negative, positive or zero.

It is important to recognizethat we are treating such a distribution as having “zero thickness”—even though, technically, there is some nonzerothickness to a typical . Provided this thickness is very small compared to the area of the surface, we canneglect it and treat the distribution as if it were truly “2D”. Surface charge density is then calculated in a mannercomparable to volume charge density. In this case, however, we measure the amount of charge found per unitarea of the surface examined.

Worked Example 2

Calculate the charge density on the surface of a conducting spherical shell of radius 0.15m when the charge on the sphere is 3.3 × 10−9 C.

Solution

Figure 3

To get thecharge density, divide the charge by the surface are of the sphere:

nC/m2

The charge density on the sphere’s surface is 11.67nC/m2.

1. If charge is spread along a line (i.e. a one-dimensional system), we have a linear charge density. Just as withsurface densities, there is no absolute requirement that the “line” be straight. We can distribute charge alongcircular arcs or even weird, squiggly lines. Linear charge densities are suitable for describing wires andthreads—essentially, any object whose cross-sectional area is tiny in comparison to its length. Linear chargedensities are represented by the symbol “λ”, and measure the amount of charge found per unit of length examined.

[Eq. 3]

Worked Example 3

A thin rod of length L = 12 cm has a charge Q = 15 nC distributed uniformly along its length. Assume that the rod lies along the x-axis, from x = 0 to x = L. Compute the charge density on this rod.

Solution

Figure 4

Charge density on rod

µC/m

Table 1 shows the different measures of charge densities we shall be using.

FIELDS OF SIMPLE CHARGE CONFIGURATIONS

In many problems it is desirable to know the distribution of the electric field and the associated potential around a charged object. For example, if the field intensity exceeds the breakdown value for the dielectric medium, sparking, or corona, can occur. From a knowledge of the field distribution, the charge surface density on conductors bounding the field and the capacitance between them can also be determined.

In the following sections the field and potential distributions for a number of simple geometries are discussed; these include the field and potential distributions around a finite line charge, infinite line charge, ring of charge, charged spheres, and charged cylinders.

Field Of A Finite Line Of Charge

Consider the field produced by a thin line of charge (see Figure 5). Let a positive charge Q be distributed uniformly as an infinitesimally thin line of length 2a with the centre at the origin, as shown in the figure. The linear charge density (charge per unit length) is then

where is in coulombs per metre when Q is in coulombs and a is in meters.

At point P on the r axis, the infinitesimal electric field dE due to an infinitesimal length of wire dz is the same as from a point charge of magnitude . Thus,

where and unit vextor in direction of l.

Since the z axis in Figure 5 is an axis of symmetry, the field has only z and r components. These are

[Eq. 4]

and

The resultant or total r component Er of the field at a point on the r axis is obtained by integrating Eq.4 over the entire line of charge. That is,

and

Figure 5. Thin line of charge of length 2a.

By symmetry the resultant z component Ez of the field at a point on the r axis is zero. Hence the total field E at points along the r axis is radial and is given by

(Eq.5)

This relation gives the field as a function of r at points on the r axis for a finite line of charge of length 2a and uniform charge density.

Worked Example 3

A charge is distributed along the z axis between  6 m with uniform charge density 25 nC/m. Calculate at a point (2,0,0) m in free space.

[Ans: 213.26 V/m]

Solution

Figure 6 for Worked Example 3.

For this problem, a = 6 m, and

Therefore

V/m

EXERCISES

1. Find at (10,0,0) due to a charge of 10 nC which is distributed uniformly along axis x between x = -5 to +5 m in free space. [Ans. V/m]
2. A line charge density 24 nC/m is located in free space on the line y =1 , z = 2.

(a)Find at P(6,-1,3).

(b)What point charge QA should be located at (-3,4,1) to cause y component of to be zero at P?

[Ans. V/m; QA = 4.43 µC]

Field Of An Infinite Line Of Charge

Consider that the line of charge in Figure 5 extends to infinity in both positive and negative z directions. By dividing the numerator and denominator of (Eq.5) by a and letting a becomes infinite the electric field intensity due to an infinite line of positive charge is found to be

(Eq. 6)

The potential difference V21 between two points at radial distances r2 and r1 from the infinite line of charge (see Figure 7) is then the work per unit charge required to transport a positive test charge from r2 to r1. Assume that r2 > r1. This potential difference is given by the line integral of Er from r2 to r1, the potential at r1 being higher than at r2 if the line of charge is positive. Thus,

or

(Eq. 8)

Figure 7. Field of an infinite line of charge

Worked Example 4

A uniform line charge of infinite length with  = 20 nC/m lies along the z axis. Find at (6, 8, 3) m.

[Ans.: V/m]

Solution

Figure 8. Diagram for Worked Example 4.

For an infinite line of charge, the electric field intensity at a radius r from the axis of the line of charge is given by the expression

Thus, for  = 20 nC/m, and

m

we obtain

Worked Example 5

On the line x = 4 and y = -4, there is uniform charge distribution with charge density  = 25 nC/m. Determine at (-2, -1, 4) m.

[Ans.: V/m]

Solution

Figure 9. Diagram for Worked Example 5.

Position vector of point A is

Position vector of point P is

Therefore, displacement vector

m

Magnitude of electric field at point P is

Worked Example 6: Field under a dc transmission line

Two long parallel conductors of a dc transmission line separated by 2 m have charges of of opposite sign. Both lines are 8 m above ground. What is the magnitude of the electric field 4 m directly below one of the wires?

Solution

Figure 10.Diagram for Worked Example 6.

where

where

V.m-1

= 1.61 kV.m-1 Ans.

Worked Example 7

A line parallel to the x-axis and passing through the points y = 1, z = 1 carries a uniform charge of 2 nC/m. Find potential at A(5, 0, 1) if

(i) V(0, 0, 0) = 0 volt

(ii) V(1, 2, 1) = 100 volts

Solution

The line is shown the figure below, which is parallel to the x axis.

Figure 10. Diagram for Worked Example 7.

Seen on the x-y plane, the locations of the given points and of the line charge are as shown in the following figure.

Figure 11. Diagram for Worked Example 7.

We have found earlier that the potential difference between two points at distances r1 and r2 from the axis of an infinite line charge is given by the expression

where, by definition,

Defining V1 as the potential at the origin and V2 as the potential at A(5,0,1), we thus have

m

m

volt (given)

and

Solving the previous equation for V2, we thus obtain

volts

Electric Field Between Two Parallel Wires

This case is of practical importance in overhead transmission lines.

With reference to Figure 12, let

D = distance between centres of the wires A and B

R = radius of each wire ( d)

Q = charge in coulomb/metre of each wire

Figure 12. Two parallel wires separated by a distance d.

Consider electric field intensity at any point P between conductors A and B. Electric field at P due to +Q coulomb/metre on A is

V/m (towards B)

Electric field intensity at P due to charge –Q coulomb/metre is

V/m (towards B)

Total electric field intensity at P,

V/m (towards B)

Hence, potential difference between the two wires is

So,

V

A plot of the electric field and voltage distributions between as a function of distance x is shown in Figure 13.

Figure 13.

Electric Field Along The Axis Of A Charged Ring

Consider a ring of uniform positive charge as shown in Figure 14. What would be the electric field at the center of the circle? What would be the electric field anywhere along the circle's axis?

Once again, westate with a smallcharge segment and add up the totalaffects of all segments atpoint P. In the following diagram, the radius of the charged ring is a, and the distance along the axis to P is x.

Figure 14. Ring of charge.

Once again, symmetry shows us that the y-components of the electric field will cancel,leaving only the x-components' contributions.

Figure 15. Ring of charge.

This derivation will be much simpler than the previous linear bisector since the values for x, a, and r keep the same value for all charge segments, q,aroundthe ring.

In order to simplify this integral wewill use the trigfunction cos q.

;

Substituting yields

and we are finished.

Worked Example 8

A charge of 2.75 μC is uniformly distributed on a ring of radius 8.5 cm. Find the electric field strength on the axis at distances of (a) 1.2 cm, (b) 3.6 cm, and (c) 4.0 m from the center of the ring. (d) Find the field strength at 4.0 m using the approximation that the ring is a point charge at the origin, and compare your results for Parts (c) and (d). Is your approximate result a good one? Explain your answer.

Solution

The magnitude of the electric field on the axis of a ring of charge is given by

where Q is the charge on the ring and a is the radius of the ring. We can use this relationship to find the electric field on the axis of the ring at the given distances from the ring.

Worked Example 9

A total charge of 40/3 nC is uniformly over a circular ring of radius 2 m placed in z = 0 plane, with centre as origin. Find the electric potential at A(0, 0, 5).

Solution

Figure 16. Diagram for Worked Example 8.

The charge is distributed over a ring hence it is a line charge. The line charge density is

The ring is shown in Figure 15.

Next, consider the differential length on the ring at point P.Amount of elemental charge over length

But , where is the incremental angle subtended by in the x-y plane measured at the origin. Therefore, incremental potential at A contributed by elemental charge dQ is

where we have used the geometric relation. Since r = 2 m, z = 5 m, and  = 1.061 nC/m, therefore

Therefore, total potential at A due to the whole ring of charge is

V

END