Electric field due to a flat hollow disk

[madafo3435]

Homework Statement

A charge distribution with a uniform positive charge surface density + σ is formed by a circular section of radii R1 and R2.

Relevant Equations

Charges to continuous charge distributions

I would like if my procedure is correct ...

Due to the symmetry of the problem, I only worry about the vertical coordinate of the field, so I will work with the magnitude of the field, and I will treat the problem in polar coordinates.

$E= \int_{R_1} ^ {R_2} \int_{0} ^ {\pi} \frac {\sigma sen(\theta)}{4\pi \epsilon _0 r^2} rd\theta dr = \frac {\sigma}{4\pi \epsilon _0} \int_{R_1}^{R_2} \frac {2}{r} dr = \frac {\sigma}{2\pi \epsilon _0} ln(\frac {R_2}{R_1})$

$\therefore \vec{E} = \frac {\sigma}{2\pi \epsilon _0} ln(\frac {R_2}{R_1}) \hat{\jmath}$
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I attach a picture of the problem below.

Is this correct?

 

[CPW]

It appears to me you are on the right track, but is the problem asking for the E field at the center of the semicircles?

Can you explain why sinθ and not cosθ?

 

[madafo3435]

It appears to me you are on the right track, but is the problem asking for the E field at the center of the semicircles?

Can you explain why sinθ and not cosθ?

Because the direction of $\vec{E}$ is given by $cos(\theta )\imath + sen(\theta )\jmath$ and due to the fixedness and uniformity, the as a component in $\imath$ total is canceled

Do you agree with me?

 

[CPW]

I agree. Nice work.
But with y-axis up in your figure, shouldn't you include a minus sign in your vector answer?

 

[madafo3435]

I agree. Nice work.
But with y-axis up in your figure, shouldn't you include a minus sign in your vector answer?

Thanks for your appreciation