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Problem 2.4 Griffiths E&M 3rd ed -- E-field above a square loop
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[QUOTE="darkexcalibur87, post: 4939986, member: 534822"] [h2]Homework Statement [/h2] 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."[h2]Homework Equations[/h2] The electric field of a line charge: [ATTACH=full]174649[/ATTACH] (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: [ATTACH=full]174650[/ATTACH] (equation 2) The problem is solved and this result is given: [ATTACH=full]174651[/ATTACH] (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.[h2]The Attempt at a Solution[/h2] 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 [ATTACH=full]201714[/ATTACH] This results in: [ATTACH=full]201715[/ATTACH] 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! [/QUOTE]
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Problem 2.4 Griffiths E&M 3rd ed -- E-field above a square loop
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