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?
Best 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.