Electric Field of a quarter circle segment

[btpolk]

Homework Statement

A quarter circle segment has a uniform linear charge density of λ. Starting with the E-field due to point charges, show that the magnitude of the E-field at the center of curvature(which is distance R away from all points on the quarter circle) is E= (kλ√(2))/R

Homework Equations

E= k∫dq/R^2 * r^

r^ is r hat

q=Rλ

k=9x10^9 or in this case just a constant

The Attempt at a Solution

I first approached this as a semi-circle and was going to divide by 2 at the end. With a semi-circle the x unit vectors I can replace r^ with y^*sinθ (didn't get the right answer so this approach is probably wrong).

E= k/R^2∫dq*r^

=((kλ)/R)*y^∫sinθ dθ

=((kλ)/R)*y^[-cos(pi)+cos(0)]

=((2kλ)/R)*y^

=((kλ)/R)*y^

I'm missing a √(2) somehow and I don't know how to get rid of the y hat.

 

[tiny-tim]

hi btpolk!

I first approached this as a semi-circle and was going to divide by 2 at the end.

well that's not going to work …

if you have E at 45° left, and E at 45° right, what does that add up to?​

(but anyway, what's so complicated about just using the correct limits of integration for a quadrant? )

 

[btpolk]

So I'm guessing I would go from 0 to pi/2? Also I'm confused about what to do with the r hat.

 

[gneill]

If you choose your angle limits carefully with respect to your axes you can take advantage of symmetry to cancel the force contributions in one axis direction for symmetric pairs of points on the circumference. Otherwise you'd have to sum the contributions for both x and y components of the force vectors (vector addition) and determine the magnitude of the resultant afterwards. That's a lot of extra work compared to making the clever choice to begin with.

 

[btpolk]

I'm having a lot of trouble visualizing this...lets see 0 to pi/4? would that cancel the x?

 

[gneill]

I'm having a lot of trouble visualizing this...lets see 0 to pi/4? would that cancel the x?

How would you choose pairs of points along that arc whose field vector x-components are equal and opposite? (It should be simple to choose such pairs if the symmetry is correct).

I suggest that you consider placing the arc symmetrically about an x-axis so that the x-axis passes through the center of the arc, the arc extending equally above and below the x-axis in the +Y and -Y directions. Then pairs of points on the arc at equal angles above and below the x-axis will be symmetric pairs. What components for the pairs cancel? What components are equal? How can you simplify the resulting math?

 

[btpolk]

I'm not sure if this is what your talking about or not but here we go:

=(k/R^2)∫dq*rhat limits being 0 to pi/4

=((kλ)/R)*xhat∫cosθ dθ

=((kλ)/R)*xhat[sin(pi/4)-sin(0)]

=((kλ√(2))/(2*R))*xhat multiply by 2 for the rest of the quarter (pi/4 to pi/2)

=((kλ√(2))/R

 

[gneill]

Well, that looks fine

 

[btpolk]

I appreciate the help gneill!

 

[tiny-tim]

hi btpolk!

(just got up :zzz: …)

actually, your semicircle method would have worked, if you'd used the correct factor …

if you have E at 45° left, and E at 45° right, that adds up to E√2 at 0°, doesn't it?

so the magnitude of the field for a quadrant is 1/√2 times that for a semicircle ​

(but of course using the correct limits is best)