# Circular arcs on net electric field

1. Sep 19, 2015

### Anon.ilbe

1. The problem statement, all variables and given/known data

http://www.webassign.net/hrw/W0490-N.jpg

The figure above shows three circular arcs centered on the origin of a coordinate system. On each arc, the uniformly distributed charge is given in terms of Q= 2*10-6C. The radii are given in terms of R = 10cm. What is the magnitude of the net electric field at the origin due to the arcs?

2. Relevant equations

dE = Kλds / r2

ds=rdθ

λ= Charge / Length ⇒ Q/2πr ⇒ Q/(2πr/4) - since only quarter is shown ⇒ 2Q/(πr)

3. The attempt at a solution

So i think I got all the steps correct except i cant figure out why i have to multiply sin(45) to get to the right answer at the end.

1. dE = Kλds / r2 - used this formula b/c it's a ring

2. dEcosθ = Kλds / r2 (cosθ) - multiplied cosθ b/c perpendicular component (sinθ) cancels each other.

3. dEcosθ = Kλrdθ / r2 (cosθ) - replaced ds to rdθ b/c cosθ is present and thus want to integrate along 90° to 180°

4. ∫dEcosθ = ∫Kλrdθ / r2 (cosθ) ⇒ E = Kλ/r ∫(cosθ)dθ - r on the numerator and r2 on the denominator cancels to make denominator just r. since r is constant, took out of the integral and just integrate cosθdθ from 90° to 180°

5. E = Kλ/r [-sin(180°)-(-sin(90°)] ⇒ E = Kλ/r - integral just turns out to be 1 and we are left with E = Kλ/r

6. E = Kλ/r - This is 1st arc so to second and third, we just replace r with 2r and 3r respectively. Then we get

E1 = Kλ/r ; E2 = Kλ/2r ; E3 = Kλ/3r

7. Replace λ with 2Q/πr (relevant equation section)

E1 = K(2Q/πr)/r ; E2 = K(2Q/πr)/2r ; E3 = K(2Q/πr)/3r

E1 = K(2Q/πr2) ; E2 = K(Q/πr2) ; E3 = K(2Q/3πr2)

8. Replace Q and r with respective charge and radius

E1 = K(2Q(R)2) ; E2 = K(-4Q(2R)2) ; E3 = K(18Q/3π(3R)2)

E1 = K(2Q(R)2) ; E2 = K(-4Q4R2) ; E3 = K(6Q(9R2)

9. Simplify

E1 = K(2Q(R)2) ; E2 = K(-QR2) ; E3 = K(2Q(3R2)

E1 = K(6Q/π3(R)2) ; E2 = K(-3Q/π3R2) ; E3 = K(2Q(3R2)

※Multiplied E1 and E2 by 3 on both numerator and denominator to simplify

E1 + E2 + E3 = Enet

Enet = K5Q/π3(R)2

--------------------------------------------------------

crap.... i dont know where I messed up.

Originally i got E1 = 2Q/4πr2ε0 + E2 = -4Q/4πr2ε0 + E3 = 6Q/4πr2ε0

Which turned out to be

Enet = Q/π2r2ε0

which turned out to be 228974

but the answer is 161903 which is just sin(45) multiplied by 228974.

why do you need to multiply sin45?

Last edited: Sep 19, 2015
2. Sep 19, 2015

### haruspex

You lost me at step 2, multiplying by cos theta. There is a valid method which does that, but does not relate to the explanation you offer. The corrsponding sin case does not cancel out (and that's where your root-2 factor would come from). Please elaborate on your explanation there.

3. Sep 20, 2015

### Anon.ilbe

Yes, I pictured the arc along the axis so the corresponding cos theta would all line up and sin theta would cancel b/c i didnt know how to incorporate cos45 and sin45 into components and calcuate it separately.

4. Sep 20, 2015

### Anon.ilbe

So i dont know if you guys can ready my terrible handwriting but here is what i worked on and at the last step i dont know why i had to multiply sin45 to get to the correct answer.

#### Attached Files:

• ###### Snapchat-2830674341006671904.jpg
File size:
22.9 KB
Views:
143
5. Sep 20, 2015

### haruspex

Maybe you need to clarify your axes. If they are the obvious ones according to the original diagram, the sines do not cancel. If you are taking the axes at 45 degrees, so that the arcs are symmetric about the x axis, the the sines do cancel, but the integration range for the angle changes (and is not what you used).

6. Sep 20, 2015

### Anon.ilbe

So if you do not line up with axis and solve it as it is, how would I have to attack the question from step 1? where you have dE = Kλds / r2
would i have to separate it into components where dE = [(Kλds / r2)(-cosθ)]i + [(Kλds / r2)(sinθ)]j
then change ds to rdθ and integrate it from 90 to 180?

7. Sep 20, 2015

### haruspex

I'm still unsure what axes you were using in your first attempt. Please clarify. What do you mean by "line up with axis"?
Yes, if you take the standard axes according to the picture, so that theta runs from 90 to 180, that is the vector form of field you need to integrate.
But, as I posted, you could just as well choose to rotate the axes 45 degrees so that you know one of the axes will have no net field. The range for theta then changes (and produces the missing sqrt(2) factor).

8. Sep 20, 2015

### Anon.ilbe

on the first attempt I was tilting axis about 45 degrees so it sine would cancel.

and "line up with axis" on the comment where I separated into components, I tried solving it like the standard problem given

the attached picture on the comment, however, i tried to solve it by moving the all three arcs with their center lined up with x-axis.

so if i did titled the axis about 45 degrees and solved this problem, the range of integration would change to 45 and 135? since you are tilting to the right 45° thus
90°-45° = 45° and 180°-45° = 135°

Last edited: Sep 20, 2015
9. Sep 21, 2015

### haruspex

Yes, try that integration range.

10. Sep 21, 2015

### Anon.ilbe

Oh my gosh! I cant believe it! Thank you so much for your guidance! I finally figured it out. Physics starting to make sense now!!

File size:
18.3 KB
Views:
153