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I have to find the direction of electric field due to this rod at point P

$dE_x =[Kdq/(d/cos\theta)^2]cos\theta$ and $dq = \lambda (d/cos\theta) d\theta$ Writing similar equations for $E_y$, then integrating these two from 0 to $pi/3$ followed by $tan^{-1}E_y/E_x$ gives me the wrong answer

I know I can get the correct answer by integrating along the length of the rod but why is using a polar system giving me the wrong answer? (I understand in this case I have chosen my intervals incorrectly since they don't even sum up to $\sqrt 3 d$).

My main question is:

Is it necessary that the intervals chosen for a Riemann sum have to be of equal size? What are the conditions/rules while choosing intervals?

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I think you have taken the dq component to be wrong.We know that $\frac{y}{d}=\tan \theta$.Therefore,$dy=d\times \sec^2 \theta \ d\theta$.In your method also,you have to divide dq by one $\cos \theta$ because you didn't take into account that $ld\theta$ where l is the hypotenuse in the big triangle does not give dy but gives $dy\cos \theta$ .Try to think of the geometry yourself.

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  • $\begingroup$ How could you prove that the perpendicular dropped from the "point of contact" of the hypotenuse with the wall onto the next line, intersects that line exactly where the arc $ld\theta$ does? Because only in that case will $dycos\theta$ become equivalent to $ld\theta$ in the limit $d\theta$ tending to zero $\endgroup$
    – xasthor
    Commented Jun 14, 2017 at 19:05
  • $\begingroup$ The two lines dropped from P to the wall will actually be parallel because the angle between them is infinitesimal and the arc $l d\(theta )$ is actually not an arc but a straight line perpendicular to the top line.Now, is the geometry clear?Anyways,did you try computing the answer with the correction factor? $\endgroup$
    – Rishabh Jain
    Commented Jun 14, 2017 at 19:57

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