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Tsunoyukami
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I'm very sorry for the initial post - I was having trouble working with LaTex (I've never used it before and the preview post wasn't showing me what I had expected it to look like - it kept giving me a square root symbol for every code).
Find the electric field a distance z above the centre of a square loop (side a) carrying uniform line charge [tex]\lambda[/tex]. (Hint: Use the result of Ex. 2.1.). [Griffths pg. 64]
Ex. 2.1 asks us 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 [tex]\lambda[/tex]." [Griffiths pg. 62]
The result of Ex 2.1 is:
[tex]\vec{E} = \frac{1}{4 \pi \epsilon_o} \frac{2L \lambda}{z} \frac{1}{\sqrt {z^{2} + L^{2}} }[/tex]
In Problem 2.4, however we have four uniform lines of charge of length a arranged such that they form a square. Supposing that the square is centred at (0, 0, 0) such that it is bounded by the lines x = -a/2, x = a/2, y= -a/2, y= a/2 (and using "the usual" orientation of the x, y and z- axis), by symmetry, the x- and y- components of the electric field at any point z along the z-axis. Therefore there exists only a z-component to the electric field.
However, in Ex. 2.1. the value of "z" is the distance from the point to the line of charge and so must be replaced by [tex]\sqrt{z^{2} + (\frac{a}{2})^{2}[/tex]. Is this correct?
Then, the square root in the result is really the separation vector between the point on the z-axis and a point along the (x) axis. So in three dimension the separation vector should be [tex]\sqrt{z^{2} + (\frac{x}{2})^{2} + (\frac{y}{2})^{2}[/tex]. Is this correct? My concern is that either x or y vary along each line of charge and I'm not sure how to account for this mathematically.I further know that we must multiply the total by 4 (since there are 4 lines of charge) and by cos[tex]\theta[/tex] (for the z-component). However, cos[tex]\theta[/tex] = [tex]\frac{z}{r}[/tex] where r is the separation vector (with the dependence on x, y and z), correct?So what I essentially expect as an answer is something along the lines of:
[tex]\vec{E_z} = \vec{E}cos\theta[/tex] = [tex]\frac{1}{4 \pi \epsilon_o} \frac{za \lambda}{\sqrt {z^{2} + (\frac{a}{2})^{2}}} \frac{4}{\sqrt{z^{2} + (\frac{x}{2})^{2} + (\frac{y}{2})^{2}}}[/tex]I'm not feeling 100% confident about this answer. I feel as if I've done most of the question properly but there's just something that doesn't feel right - mostly the dependence on x and y - the field shouldn't (at least as far as I can tell) depend on the value of x or y - so I think I might need to rewrite them in terms of a somehow? Thanks for your patience while I struggled with Latex and I appreciate any and all responses to my questions here. Hopefully I'm on the right track, but if anyone is willing to help out that would be great!
Find the electric field a distance z above the centre of a square loop (side a) carrying uniform line charge [tex]\lambda[/tex]. (Hint: Use the result of Ex. 2.1.). [Griffths pg. 64]
Ex. 2.1 asks us 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 [tex]\lambda[/tex]." [Griffiths pg. 62]
The result of Ex 2.1 is:
[tex]\vec{E} = \frac{1}{4 \pi \epsilon_o} \frac{2L \lambda}{z} \frac{1}{\sqrt {z^{2} + L^{2}} }[/tex]
In Problem 2.4, however we have four uniform lines of charge of length a arranged such that they form a square. Supposing that the square is centred at (0, 0, 0) such that it is bounded by the lines x = -a/2, x = a/2, y= -a/2, y= a/2 (and using "the usual" orientation of the x, y and z- axis), by symmetry, the x- and y- components of the electric field at any point z along the z-axis. Therefore there exists only a z-component to the electric field.
However, in Ex. 2.1. the value of "z" is the distance from the point to the line of charge and so must be replaced by [tex]\sqrt{z^{2} + (\frac{a}{2})^{2}[/tex]. Is this correct?
Then, the square root in the result is really the separation vector between the point on the z-axis and a point along the (x) axis. So in three dimension the separation vector should be [tex]\sqrt{z^{2} + (\frac{x}{2})^{2} + (\frac{y}{2})^{2}[/tex]. Is this correct? My concern is that either x or y vary along each line of charge and I'm not sure how to account for this mathematically.I further know that we must multiply the total by 4 (since there are 4 lines of charge) and by cos[tex]\theta[/tex] (for the z-component). However, cos[tex]\theta[/tex] = [tex]\frac{z}{r}[/tex] where r is the separation vector (with the dependence on x, y and z), correct?So what I essentially expect as an answer is something along the lines of:
[tex]\vec{E_z} = \vec{E}cos\theta[/tex] = [tex]\frac{1}{4 \pi \epsilon_o} \frac{za \lambda}{\sqrt {z^{2} + (\frac{a}{2})^{2}}} \frac{4}{\sqrt{z^{2} + (\frac{x}{2})^{2} + (\frac{y}{2})^{2}}}[/tex]I'm not feeling 100% confident about this answer. I feel as if I've done most of the question properly but there's just something that doesn't feel right - mostly the dependence on x and y - the field shouldn't (at least as far as I can tell) depend on the value of x or y - so I think I might need to rewrite them in terms of a somehow? Thanks for your patience while I struggled with Latex and I appreciate any and all responses to my questions here. Hopefully I'm on the right track, but if anyone is willing to help out that would be great!
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