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Example: Two Segments of Charge
Suppose we have two segments of uniformly distributed charge, one with total charge
Facts
- One segment lies on the
-axis stretching from to , with charge uniformly distributed. - The other segment lies on the
-axis stretching from to , with charge uniformly distributed. - The point
is at the arbitrary location - The electric field due to a point charge is
- The electric field at
is the superposition of contributions from the two segments:
Goal
- Find
.
Representations
Solution
Approximation
We begin with an approximation, which will make our calculations simpler, and makes sense based on our representation:
- The thicknesses of both segments are infinitesimally small, and we can approximate them as line segments.
This example is complicated enough that it's worthwhile to make a plan.
Plan
We will use integration to find the electric field from each segment, and then add the electric fields together using superposition. We'll go through the following steps.
- For the first segment, find the linear charge density,
. - Use
to write an expression for . - Assign a variable location to the
piece, and then use that location to find the separation vector, . - Write an expression for
. - Figure out the bounds of the integral, and integrate to find electric field at
. - Repeat the above steps for the other segment of charge.
- Add the two fields together to find the total electric field at
.
Because we know that electric fields add through superposition, we can treat each of the charges separately, find the electric field, then add the fields together at











Assumption
The charge is evenly distributed along each segment of charge. This allows each little piece of charge to have the same value for charge density along each line.
The separation vector






































Now, we have enough to define the electric field from the small piece (

























































Next, we integrate over the entire segment to find an expression for its contribution to the electric field vector at

















































Next, we can do a similar analysis to find the electric field vector contribution from the segment that lies along the
From here, we can find


























































To find the contribution from the entire segment, we again must determine the endpoints of our integration. Our variable of integration is


















































Then the final electric field vector at
























































































































































































