# Calculating Electric Fields of Charge Distributions

#### 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 1.5.1.

(Figure 1.5.1)

**Figure 1.5.1**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:

- charge per unit length (
**linear charge density**); units are coulombs per metre () - charge per unit area (
**surface charge density**); units are coulombs per square metre () - charge per unit volume (
**volume charge density**); units are coulombs per cubic metre ()

Then, for a line charge, a surface charge, and a volume charge, the summation in Equation 1.4.2 becomes an integral and

is replaced by

,

, or

respectively:

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

,

, or

as the case may be, expressed in terms of

, and also expressing the charge density function appropriately. It may be constant; it might be dependent on location.

Note carefully the meaning of

in these equations: It is the distance from the charge element

to the location of interest,

(the point in space where you want to determine the field). However, don’t confuse this with the meaning of

; we are using it and the vector notation

to write three integrals at once. That is, Equation 1.5.2 is actually

#### EXAMPLE 1.5.1

#### Electric Field of a Line Segment

Find the electric field a distance

above the midpoint of a straight line segment of length

that carries a uniform line charge density

.

#### Strategy

Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces of length

, each of which carries a differential amount of charge

. 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 1.5.2). 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.

(Figure 1.5.2)

**Figure 1.5.2**A uniformly charged segment of wire. The electric field at point*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

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

The total field

is the vector sum of the fields from each of the two charge elements (call them

and

, for now):

Because the two charge elements are identical and are the same distance away from the point

where we want to calculate the field,

, so those components cancel.

This leaves

These components are also equal, so we have

where our differential line element

is

, in this example, since we are integrating along a line of charge that lies on the

-axis. (The limits of integration are

to

, not

to

, because we have constructed the net field from two differential pieces of charge

. If we integrated along the entire length, we would pick up an erroneous factor of

.)

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

and

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

and

Substituting, we obtain

which simplifies to

#### 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 1.4

How would the strategy used above change to calculate the electric field at a point a distance

above one end of the finite line segment?

#### EXAMPLE 1.5.2

#### Electric Field of an Infinite Line of Charge

Find the electric field a distance

above the midpoint of an infinite line of charge that carries a uniform line charge density

.

#### 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

where our differential line element

is

, in this example, since we are integrating along a line of charge that lies on the

-axis. Again,

Substituting, we obtain

which simplifies to

#### 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

,

dominates the

in the denominator, so that Equation 1.5.5 simplifies to

If you recall that

, the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected.

In the limit

, 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:

(1.5.6)

An interesting artifact of this infinite limit is that we have lost the usual

dependence that we are used to. This will become even more intriguing in the case of an infinite plane.

#### EXAMPLE 1.5.3

#### Electric Field due to a Ring of Charge

A ring has a uniform charge density

, 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 1.5.3.

(Figure 1.5.3)