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A half-circular segment with radius $R$ has total charge $Q$ distributed uniformly along length $L$. The total electric field at the origin due to this charge segment is $E_0 \hat{i}$. What is the electric field $E$ at the origin produced by a quarter-circular segment with the same radius $R$ and total charge $Q$?

enter image description here

My thinking was that now there is no symmetry cancelling the $j$ component and also, given the direction of the vectors, that the $j$ would be positive and the $i$ would be negative (this all seems correct).

However, I derived my E-field formula for the half segment to be $$E=2KQ/\pi^2 \, .$$ Therefore, I assume $r=\sqrt 2$, $E=E_0 \hat{i}$. Given this, I said that for the quarter segment, $E=KQ/\pi r^2$, meaning for the same $KQ/r^2$, I would have $2E_0 \hat{i}$ and $2E_0 \hat{j}$, giving $[-2E_0 \hat{i}+2E_0 \hat{j}]$.

However, the correct answer is in fact $E_0 \hat{i} + E_0 \hat{j}$. Why?

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    $\begingroup$ Welcome to Physics Stack Exchange. This post would be a lot easier to understand if it had a diagram. If you add one, you're more likely to get a good answer. Also, note that I went through and fixed all the math formatting. Please do this in your future posts, using our guide on mathjax. $\endgroup$ – DanielSank Feb 13 '17 at 19:01
  • $\begingroup$ Thanks, I am very new so didn't realize that these were options. Ill have a look at the guide for future as it would also be much easier for me too! $\endgroup$ – David Rowlands Feb 13 '17 at 19:02
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This diagram should make it clear:

enter image description here

The blue line is the vector sum of the red and green vectors - which are generated by the quarter segments of charge respectively. The vertical components cancel, the horizontal components add. So the horizontal (or vertical) component of either of them is $\frac{E_0}{2}$

Now if you have a charge Q on a quarter segment, the charge density is twice as big as if the charge was distributed over the semicircle

I think you should be able to see how to scale things from here.

Obviously - the signs of things will depend on the exact geometry. Without a diagram from you I didn't want to attempt to get into that; I am just trying to explain the factor 2.

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  • $\begingroup$ Perhaps I am misreading this, but your explanation seems to support my thoughts that it should equal $[-2E_0 \hat{i}+2E_0 \hat{j}]$ since the density is twice as large. However, the correct answer is $[-E_0 \hat{i}+E_0 \hat{j}]$. $\endgroup$ – David Rowlands Feb 13 '17 at 19:57
  • $\begingroup$ The green segment by itself (with charge Q/2) would give a field of $E_0/2$ in X and Y. Double the charge density and it produces the field given in the answer. $\endgroup$ – Floris Feb 13 '17 at 20:59

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