Why is dq = λR dθ in electric field integration for a charged arc?

[yeahhyeahyeah]

Homework Statement

In the picture given to me, 3/4 of a circle is drawn around the origin. Basically every quadrant but the first. The radius is R and the charge density is λ. It says, find the x-component of the electric field on a point charge at the origin.

Homework Equations

I integrate coulomb's law so I get
∫kdq/r^2cos(Θ) where Θ goes from pi/2 to pi, because the bottom two quadrants cancel each other out

Now, the solutions given say to do
dq = λR

why do I replace dq with λR ?? I'm very confused about what I'm doing

if q = λ2piR
shouldn't dq= λ ds where ds goes from 0 to 2piR, so why the heck does that give me the wrong answer, why is it just dq = λR, I don't understand why the charge density can just be multiplied by the radius, what does that give you? Shouldn't it be multiplied by the actual length of that bit, as in λ2piR or however much of the circumfrence is being used?

 

[Carid]

because the bottom two quadrants cancel each other out

Wouldn't it be rather the second and fourth quadrants who cancel each other out?

 

[yeahhyeahyeah]

Err... well I think if I was looking at both x and y components then yes, but as I'm only finding the x component of the E field it doesn't really matter, either of the quadrants on the left can cancel out the one on the right, but you're right, the y's of the fourth are canceled by the y's of the secnod

 

[Redbelly98]

Now, the solutions given say to do
dq = λR

Well, that's not quite right. Almost, but not completely right.

why do I replace dq with λR ?? I'm very confused about what I'm doing

if q = λ2piR
shouldn't dq= λ ds

dq= λ ds is correct.
Now, can you related "ds" to R and something to do with the angle θ? Drawing a figure with ds would probably help with this.

where ds goes from 0 to 2piR, so why the heck does that give me the wrong answer, why is it just dq = λR, I don't understand why the charge density can just be multiplied by the radius, what does that give you? Shouldn't it be multiplied by the actual length of that bit, as in λ2piR or however much of the circumfrence is being used?

Yes, λ should be multiplied by the length of the bit, "ds". See my hint/question above.