Problem 2.4 Griffiths E&M 3rd ed -- E-field above a square loop

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1. Dec 9, 2014

darkexcalibur87

1. The problem statement, all variables and given/known data
The problem states, "Find the electric field a distance z above the center of a square loop (sides of length a) carrying a uniform line charge λ. " The hint says to use the result of example 2.1.

Example 2.1 is a similar problem, but instead of a square loop you are asked to "Find the electric field a distance z above the midpoint of a straight line segment of length 2L which carries a uniform line charge λ."

2. Relevant equations
The electric field of a line charge: (equation 1)

The example starts by putting this equation (in the form of dE) into the z-direction by putting a cosθ at the end. It looks like this: (equation 2)

The problem is solved and this result is given: (equation 3)
The example states that this equation is in the z-direction, and this is the equation that will be used for problem 2.4.

3. The attempt at a solution
I am following the solutions manual on this one, as I am doing all of this for self-study. There is just one thing that I cannot understand from the solution.

To adapt the above equation to problem 2.4, we make the substitution that L=a/2 , or L=2a.
Then substitute

This results in:

Then multiply by 4 because we have a square loop instead of just one line segment.

Here is what I do not understand: the solutions manual states to now multiply by cosθ to get only the z-direction. But wasn't that already done to derive equation 3 that we used for this problem? The book even states that this equation is in the z-direction. So why do it again??

I hope this is enough information, thanks for your help!

2. Dec 10, 2014

TSny

Hello, and welcome to PF!

Think about the direction of the electric field due to one side of the square at the point that is located a distance z above the midpoint of the square.

Aside: Logically speaking, the solution manual should have first multiplied by $\cos \theta$ and then multiplied by 4.