# Electric field of a half and quarter segment

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 $$Q$$?

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?

• 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. Feb 13, 2017 at 19:01
• 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! Feb 13, 2017 at 19:02

## 1 Answer

This diagram should make it clear:

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.

• 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 $[-E_0 \hat{i}+E_0 \hat{j}]$. Feb 13, 2017 at 19:57
• 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. Feb 13, 2017 at 20:59