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.

Take-Home Component

A similar segment of charge - Q is joined to the segment of + Q as shown below.

dipolar line of charge

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


Solution for Takehome Component:

The solution for the left-hand side of the charge distribution, the positively-charged part was solved as the in-class part of this quiz. (See solution.)

Electric field due to + charge

pointing to the right.

The negatively-charged part will be considered separately and then added to the result above invoking the superposition principle.

Step 1: Understand the geometry

Same as positively-charged part.

Step 2: Span the charge distribution

Same as positively-charged part, except that the integration variable x is defined to be positive to the right as shown. The entire charge distribution may again be spanned by varying x from 0 to a. Alternatively, the variable u with an origin at the point of evaluation P may 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 again be defined as Q / a. The negative sign of the charge will be handled here via the direction of the electric field contribution and not by defining a negative charge density.

Step 3: Evaluate the contribution from the infinitesimal charge element

electric field infinitesimal

The infinitesimal contribution to the electric field is

electric field infinitesimal, ready to integrate

Step 4: Exploit symmetry as appropriate

Nothing to do...

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!

Again, for r >> a this expression reduces to the expected result for a point charge Q. Keep in mind that the charge is now negative so this electric field points to the left, rather than to the right as that from the positively-charge piece.

Step 8: Superposition

E- is larger than E+, as expected since the negative charge is closer to the point if evaluation. The magnitude of the net electric field will thus be E- - E+, to the left.

Superposition of two solutions

Step 9: Check Far-Field Limit

As expressed above, the limit may be checked without the binomial theorem. As a becomes small compared to r, the denominator becomes r3 and the result behaves as a point dipole of dipole moment p = Qa, the charge times the "center of mass" separation.

Note that the binomial theorem could be applied with success to the second line above, as demonstrated here.


"http://www.physics.udel.edu/~watson/phys208/quiz3soln.html"
Last updated Mar. 11, 1998.
Copyright George Watson, Univ. of Delaware, 1998.