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## Homework Statement

Suppose the charge Q on the ring of Fig. 28 was all distributed uniformly on only the upper half of the ring, and no charge was on the lower half. Determine the electric field [itex]\vec {E}[/itex] at P.

## Homework Equations

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[tex] E = k\frac {q} {r^2} [/tex]

## The Attempt at a Solution

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Starting with the charge density,

[tex]\lambda = \frac{Q}{\pi a} = \frac{dq}{dl}[/tex]

electric field for dq

[tex] dE = k\frac{dq}{x^2+a^2}[/tex]

Let's find its x component

[tex]dE_x = dEcos\theta = k\frac{dq}{x^2+a^2}cos\theta = kx\frac{dq}{(x^2+a^2)^{3/2}}[/tex]

[tex]dE_x = \frac{k\lambda x}{(x^2+a^2)^{3/2}}dl[/tex]

Wıth the limits being [itex]0[/itex] and [itex]\pi a[/itex]

[tex]E_x = \frac{k\lambda x \pi a}{(x^2+a^2)^{3/2}} = \frac{kqx}{(x^2+a^2)^{3/2}}[/tex]

This is

**correct**according to my textbook.

**However, I end up with a wrong y component when I apply the same logic for the vertical.**

[tex]E_y = -\frac{k\lambda a \pi a}{(x^2+a^2)^{3/2}} = -\frac{kqa}{(x^2+a^2)^{3/2}}[/tex]

But the correct answer for the vertical is

[tex]E_y = -\frac{kq2a/\pi}{(x^2+a^2)^{3/2}}[/tex]

Having searched the internet for this question, I found that the solution for the y axis includes an arc length element, which is something like this.

[tex]\lambda = \frac {dq} {a d\theta}[/tex]

Then I suppose, it integrates from [itex]0[/itex] to [itex]\pi[/itex] for a semi circle.

So my question is, how does that differ if one uses [itex]d\theta[/itex] or [itex]dl[/itex] for integration? Also, why did I get the x component correct, even though the fact that I used [itex]dl[/itex]?