PHYS208 Fundamentals of Physics II

Quiz 3 -- Charge Distributions

Charge + Q is distributed uniformly along a line segment of length a.

line of charge

  1. Derive an expression for the electric field at a point P a distance r from the end of the segment.
  2. Demonstrate that for points far away from the segment, this expression for the electric field behaves like that of a point charge.


Step 1: Understand the geometry

This configuration is similar to the line segment of charge in HRW exercises 23-33P and 25-42P. The difference from 25-42P is that the charge is distributed uniformly and we seek to find the electric field rather than the electric potential; in 23-33P you were asked to find the electric field at a different location relative to the charge distribtuion.

Step 2: Span the charge distribution

Since the charge is on a line segment and the point of evaluation is on its axis, the translation variable x with the origin at the end is a convenient choice. The entire charge distribution can then be spanned by varying x from 0 to a. Alternatively, the variable u with an origin at the point of evaluation P could be selected, with limits of integration from r to r + a.

definition of variable of integration

Assuming a uniform charge distribution about the ring, the linear charge density lambda will be Q / a. In terms of the charge density, the infinitesimal charge element will be

charge infinitesimal

Step 3: Evaluate the contribution from the infinitesimal charge element

electric field infinitesimal

Do not use this formula "blindly"; the variable r is a placeholder for the distance separating the charge infinitesimal and the point of evaluation. Referring to the diagram below, that distance is r + x.

electric field infinitesimal

Thus the infinitesimal contribution to the electric field is

electric field infinitesimal, ready to integrate

Step 4: Exploit symmetry as appropriate

Since the contribution to the electric field from each charge infinitesimal points in the same direction, to the right, there is no further simplication of the problem added by symmetry arguments.

Step 5: Set up the integral

integral setup for the electric field

Substitution of variable from x to u = r + x has been made to facilitate solution of the integral.

Step 6: Solve the integral

solution of the integral for the electric field

Step 7: The result!

For r >> a this expression reduces to the expected result for a point charge Q.

Last updated Mar. 7, 1998.
Copyright George Watson, Univ. of Delaware, 1998.