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# CLFds2

Module by: George Brown. E-mail the author

Summary: This is Page 2 of a PLTL workshop for introductory physics, at the undergraduate sophomore level. It may also be used in Workshop Physics, or simply as supplemental material for the course.

Coulomb Law Fields - Page 2 of 6

## Electric Field due to Multiple Point Source Charges

Suppose there are more than one point source charges contributing to the electric field. A fairly simple example with two source charges shown in green (this group formed by positive and negative charges of equal magnitude is called a dipole) and two different field points (black dots), is diagramed in Figure 1. We assume that the source charges are fixed in space, separated by the distance aa, and calculate the electric field at each of the two field points.

Note the coordinate system that has been chosen. The origin is placed at the midpoint between the two source charges, and the vertical axis, coordinate zz, is aligned along the source points. The diagram is shown in the z xz x plane. The field points chosen lie on the zz and xx axes.

The total field on any field point is due to applications of Equation 1 for each of the source charges. So the total field is the sum of these individual fields. (Like the Coulomb forces, the Coulomb electric fields superpose.) So, for any group of NN point source charges q n q n , the electric field at any field point r r is given by

### Equation 2

E [ r ] = 1 4 π ϵ 0 n = 1 N q n r n 2 r ^ n E [ r ] = 1 4 π ϵ 0 n = 1 N q n r n 2 r ^ n
(1)

Remember that the meaning of the vector r n = r n r ^ n r n = r n r ^ n is the vector from the n th n th source point to the field point. This emphasizes the important conceptual point that Equation 2 is a sum over the source point charges. The field points enter the sum only in an indirect way, through the determination of the values of the r n r n . The geometry of the source points and the field points decides the values of the r n r n used in any specific calculation.

In this example we explicitly apply Equation 2 to each of the chosen field points for the dipole source.

First we apply Equation 2 to our dipole and the field point at r 1 = { 0 , 2 a } = 2 a z ^ r 1 = { 0 , 2 a } = 2 a z ^ . We use a subscript "1" to refer to this field point, and subscripts "+" and "-" to refer to the positive and negative source charges, respectively. Then r + 1 = 3 a 2 z ^ r + 1 = 3 a 2 z ^ and r - 1 = 5 a 2 z ^ r - 1 = 5 a 2 z ^ . These are the results from the geometry of the field point relative to the source points. Using these results in Equation 2 yields E [ r 1 ] = 1 4 πϵ 0 ( q z ^ ( 3 a 2 ) 2 - q z ^ ( 5 a 2 ) 2 ) E [ r 1 ] = 1 4 πϵ 0 ( q z ^ ( 3 a 2 ) 2 - q z ^ ( 5 a 2 ) 2 ). Simplifying this gives the solution E [ r 1 ] = 16 q 225 π ϵ 0 a 2 z ^ E [ r 1 ] = 16 q 225 π ϵ 0 a 2 z ^ . (Don't believe what you read. If it appears to you that some steps were left out in reaching this solution, then you supply the intermediate steps yourself to verify the result.)

Next we apply Equation 2 to find the field at the second field point, located at r 2 = { 2 a , 0 } = 2 a x ^ r 2 = { 2 a , 0 } = 2 a x ^ . The source point locations are r ± = z ^ ( ± a 2 ) r ± = z ^ ( ± a 2 ) , and r ± + r ± 2 = r 2 r ± + r ± 2 = r 2 . A diagram of these vectors might be helpful (see Figure 2).

Thus the geometry gives us the vectors we need to substitute into Equation 2: r + 2 = a ( 2 x ^ - z ^ 2 ) r + 2 = a ( 2 x ^ - z ^ 2 ) and r - 2 = a ( 2 x ^ + z ^ 2 ) r - 2 = a ( 2 x ^ + z ^ 2 ) . Note that r + 2 2 = ( r + ) 2 · ( r + ) 2 = 17 4 a 2 = r - 2 2 r + 2 2 = ( r + ) 2 · ( r + ) 2 = 17 4 a 2 = r - 2 2 , and r ^ ± 2 = ( 4 x ^ z ^ ) 17 r ^ ± 2 = ( 4 x ^ z ^ ) 17 .

Substituting the geometry results into Equation 2 gives us the statement that E [ r 2 ] = 1 4 π ϵ 0 ( 4 q ( 4 x ^ - z ^ ) 17 17 a 2 - 4 q ( 4 x ^ + z ^ ) 17 17 a 2 ) E [ r 2 ] = 1 4 π ϵ 0 ( 4 q ( 4 x ^ - z ^ ) 17 17 a 2 - 4 q ( 4 x ^ + z ^ ) 17 17 a 2 ). Simplifying this, we find our solution is E [ r 2 ] = ( 2 q ) ( - z ^ ) 17 17 π ϵ 0 a 2 E [ r 2 ] = ( 2 q ) ( - z ^ ) 17 17 π ϵ 0 a 2 .

## Coulomb Fields of Extended Charged Objects

For sources that are extended bodies rather than a collection of isolated points, we need a more sophisticated manner of adding up the contributions of single charges to get the total electric field at a field point. This summing procedure is provided by the integral calculus. Basically, the vector sum of Equation 2 ( E [ r ] = 1 4 π ϵ 0 n = 1 N q n r n 2 r ^ n E [ r ] = 1 4 π ϵ 0 n = 1 N q n r n 2 r ^ n ) must be replaced by an integration.

The concept involved is illustrated in the diagram below. This shows an extended charged three-dimensional cylinder, with the origin at the center of the cylinder, and a field point on the axis outside the cylinder. The cylinder is shown as a wireframe, with all but one infinitesimal element of it, labeled q q , stripped away so that we can apply Equation 1 to that element. But the idea is that the cylinder is filled with such elements of charge, and we need to add all up all their contributions to find the total electric field at the field point.

The source shown is composed of a right circular cylinder of charge, and a field point is located on the axis of the cylinder. An infinitesimal element of the source is designated by q = ρ V q = ρ V

So we wish to apply Equation 1 to the relationship between the element of charge q q and the field point at r r . Then we sum the result over all the charge elements q q in the volume V V of the cylinder.

Starting with the application of Equation 1 ( E Q [ r ] = Q 4 π ϵ 0 r 2 r ^ E Q [ r ] = Q 4 π ϵ 0 r 2 r ^ ) to the charge element q q , we have an element of the electric field at the field point:

E [ r ] = q 4 π ϵ 0 η 2 η ^ E [ r ] = q 4 π ϵ 0 η 2 η ^
(2)

Note that it is the vector η = r - r η = r - r that appears here, not the vector r r . The vector involved in a Coulomb calculation is always the vector from the source point to the field point. Here, we use r r exclusively to represent the vector from the origin to the field point. Because they have different meanings, we use different symbols to represent these two vectors. Please note this carefully, because textbooks often are not careful to make this distinction clear, resulting in a great deal of confusion for students.

(Most of the higher-level textbooks simply use r - r r - r everywhere, instead of defining another symbol to represent this vector. This has the advantage of avoiding misunderstanding, but makes writing this basic relationship ugly and awkward. Compare the form given above to E [ r ] = q ( r - r ) 4 π ϵ 0 ( r - r ) 3 E [ r ] = q ( r - r ) 4 π ϵ 0 ( r - r ) 3 . The two forms are equivalent, but which would you rather write and think about?)

Let's represent the total charge on the cylinder by Q Q. Then the total electric field is obtained by integrating the effect of each and every charge element q q at the field point. We can represent this by

E [ r ] = 1 4 π ϵ 0 Q η ^ η 2 q E [ r ] = 1 4 π ϵ 0 Q η ^ η 2 q
(3)

Note that η ^ η 2 η ^ η 2 cannot come outside the integration; this quantity changes because r r changes as q q is swept through Q Q. (Remember that η = r - r η = r - r .) And what is the meaning of this integral? The integration is over a charge variable, but it's mixed up in the integrand with the space variable η η . In the calculus classes, you learn to integrate over space variables; can this integral be recast as an integration over spatial variables?

The trick is to use a new variable that ties the local charge to the local space. It is called the volume charge density, defined as ρ := q V ρ := q V , where V V is an element of the charged volume (we use the prime to indicate spatial quantities specific to sources). Using this, we can make the replacement q = ρ V q = ρ V , and our basic equation becomes

### Equation 3

E [ r ] = 1 4 π ϵ 0 V ρ η ^ η 2 V E [ r ] = 1 4 π ϵ 0 V ρ η ^ η 2 V
(4)

Now we have an integration purely over spatial variables. This Equation 3 is the basic Coulomb law for the electrostatic field of an extended charged object. All of the local charge information is wrapped up in the charge density variable ρ ρ. The integration occurs as r r sweeps out the volume V V of the charged object. In many cases ρ ρ may depend on r r .

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