22.11:
Electric Field of a Continuous Line Charge
Let a rod of length L carry a uniform line charge density, λ. What is the electric field at a perpendicular distance d from its midpoint? It is the vector integral of the electric field due to infinitesimal line elements.
Choose the origin at the midpoint of the rod. Consider the electric field due to a line element at point x along the rod, which has a mirror element at –x. The x-components of their fields at P exactly cancel. Since this is true for each pair, the resultant field is perpendicular to the rod.
The perpendicular components of each pair reinforce. Since they are of the same magnitude, the resultant field is twice the integral over half of the rod, which depends on its total charge q, L, and d.
At large distances, the expression reduces to the field of a point charge, indicating that the line element is too small for its distribution to be significant.
At small distances, the expression reduces to one that falls inversely with the distance.
22.11:
Electric Field of a Continuous Line Charge
In physics, symmetry in a system means that something in the considered system remains unchanged due to a specific operation to which it is subjected. For example, consider a horizontal square. The square looks the same if its right and left sides are interchanged. Hence, it is symmetric under a right-left interchange.
In calculations of electric fields, symmetry is of great use. For example, while calculating electric fields of continuous charge distributions.
Consider a line element with a uniform line charge density. The electric field of interest is at a point perpendicular to the line and exactly along its midpoint.
First, note that the principle of superposition allows the electric field to be written as a vector integral along the rod. The objects being integrated are infinitesimal vectors, each possessing a magnitude and a direction.
Next, notice that the rod is symmetric on the right and left of the perpendicular line going through its midpoint. This symmetry is then exploited to ease the calculation. Resolve the electric field caused by any line element into two components: one parallel to the rod and the other perpendicular to it. Notice that for each such line element, there is one exactly on the other side of the midpoint, creating an electric field whose parallel component is equal in magnitude and opposite in direction. Since this symmetry holds for each such pair along the rod, the net sum of the parallel components vanishes, leaving the perpendicular component.
Notice that the same symmetry ensures that the perpendicular components of the pair's electric field are equal in magnitude and direction, thus reinforcing each other. Hence, the total electric field is twice the vector sum of the perpendicular components of any one half of the rod.
If the field point is far from the rod, the expectation is that the rod's internal distribution should not matter, and the field resembles that of a point charge. That is what happens under the large distance approximation.
The small distance approximation gives rise to an interesting result. The electric field points away from the rod and falls off as the first power of the distance. Thus, there are two extremes. In one, the electric field falls off as the square of the distance. On the other, it falls as the distance.
Suggested Reading
- OpenStax. (2019). University Physics Vol. 2. [Web version]. Retrieved from 5.5 Calculating Electric Fields of Charge Distributions – University Physics Volume 2 | OpenStax; section 5.5; page 206.