# Coulomb's Law on a line again.

1. Aug 29, 2009

### exitwound

Coulomb's Law on a line....again.

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

2. Relevant equations

$$E=\frac{kQ}{d^2}$$

3. The attempt at a solution

I've been working on this one for three days now and can't come up with a valid answer.

I've broken this problem up into two sections, one for the upper ring segment, and one for the bottom ring segment.

Upper Segment

$$E=\frac{kQ}{d^2}$$ \\Coulomb's Law
$$dQ=\lambda ds$$ \\charge on the differential
$$ds= R d\Theta$$ \\area of the differential

$$dE=\frac{kdQ}{d^2}$$
$$dE=\frac{k\lambda R d\Theta}{R^2}$$

$$E=\int_0^{\pi /2}dE$$ \\Electric field is the sum of all the differential electric fields.

$$E=\int_0^{\pi /2}\frac{k\lambda R d\Theta}{R^2}$$

$$E=\frac{k\lambda R}{R^2}\int_0^{\pi /2}d\Theta$$

$$E=\frac{k\lambda}{R}\int_0^{\pi /2}d\Theta$$

$$E=\frac{k\lambda}{R}(\pi/2 - 0)$$

$$E=\frac{k\lambda \pi}{2R}$$

$$\lambda = \frac{Q}{L}$$
$$\lambda = \frac{Q}{\frac{R\pi}{2}}$$

$$E=\frac{2kQ \pi}{2\pi R^2}$$

$$E=\frac{kQ}{R^2}$$

$$E=\frac{(9E9)(5.5E^-12)}{(.0150)^2}=220 N/C$$

This is the Electric field generated by the upper segmant of the ring. The angle at which it points is 45 degrees below horizon, or 315 degrees.

Lower Segment

The lower segment is identical. The same charge magnitude exists on the ring. The distance is the same to the ring. The only difference is the direction of the resulting force because the charge on the ring is negative. Therefore, the Electric field is also 220 N/C but its direction is towards the ring, or 135 degrees.

Both Segments

The resulting forces added together produce a direction directly down, or 270 degrees from horizontal. Since we have canceling x-components of the two forces, the y-components are all that matter.

$$E_y=Esin(45) = 220 sin(45) = 155.56$$ \\Ey component of upper half.
$$E_y=Esin(45) = 220 sin(45) = 155.56$$ \\Ey component of lower half.

Added together, is 311.13 N\C in the 270 degree direction.

What am I doing wrong??

2. Aug 29, 2009

### gabbagabbahey

Re: Coulomb's Law on a line....again.

Well, you have $dE=\frac{\lambda k d\theta}{R}$, but just like the total field, each infinitesimal portion of the field will be a vector, so you need to find $d\textbf{E}$ and integrate that in order to find $\textbf{E}$....you will find that the x-component integrates to zero, consistent with your answer to (b) and the y component integrates to something negative

3. Aug 29, 2009

### exitwound

Re: Coulomb's Law on a line....again.

so you're saying I need to do this?

$$dE=\frac{\lambda k cos \Theta d\Theta}{R}$$

4. Aug 29, 2009

### gabbagabbahey

Re: Coulomb's Law on a line....again.

Sort of, $dE\cos\theta$ will give you a component of $d\textbf{E}$, but which component (+x,+y,-x,or -y component)?

5. Aug 29, 2009

### exitwound

Re: Coulomb's Law on a line....again.

I'm lost then. I have no idea.

6. Aug 29, 2009

### gabbagabbahey

Re: Coulomb's Law on a line....again.

Color in a small section of your wire (in the upper region) and call that section $dQ$, label the angle $\theta$ however you want, and draw a vector from $dQ$ to $P$....Colulomb's law tells you that the field $d\textbf{E}$ due to $dQ$ points along that vector right? What component of that will point in the x-direction? How about the y-direction?

7. Aug 29, 2009

### exitwound

Re: Coulomb's Law on a line....again.

This is what I've been doing. You should see my pages and pages of pictures.

Upper section:
(I'm taking Theta to be from the +y axis towards the -x axis)

E points in the 315 degree direction.
The y-component of that would be Ecos(Theta).
The x-component of that would be Esin(Theta).

8. Aug 29, 2009

### gabbagabbahey

Re: Coulomb's Law on a line....again.

http://img199.imageshack.us/img199/7655/efield.th.jpg [Broken]

You should see clearly, that the x-component of $d\text{E}$ is $dE_x=dE\sin\theta$ while the y-component is $dE_y=-dE\cos\theta$.....right?

Last edited by a moderator: May 4, 2017
9. Aug 29, 2009

### exitwound

Re: Coulomb's Law on a line....again.

Indeed. I left the negative off the cos because I knew the direction. My picture looks identical to yours.

10. Aug 29, 2009

### gabbagabbahey

Re: Coulomb's Law on a line....again.

Okay, so integrate now.....what do you get for the components of the field due to the upper section of wire?

11. Aug 29, 2009

### exitwound

Re: Coulomb's Law on a line....again.

$$E=\int_0^{\pi/2}\frac{k\lambda (-cos\Theta) d\Theta}{R}$$

$$E=\frac{-k\lambda}{R}\int_0^{\pi/2}cos\Theta d\Theta$$

$$E=\frac{-k\lambda}{R}(sin\Theta|_0^{\pi/2})$$

$$E=\frac{-k\lambda}{R}(1-0)$$

$$E=\frac{-k\lambda}{R}$$

12. Aug 29, 2009

### gabbagabbahey

Re: Coulomb's Law on a line....again.

Well that's $E_y$...you know from symmetry that $E_x$ from the upper section should cancel $E_x$ from the lower section, so you don't have to calculate each $E_x$ but you probably shuld anyways, so that you can check that the two do in fact cancel (if they don't then you know you've done something wrong)

Anyway, what about $E_y$ from the lower segment....what do you get for that?

13. Aug 29, 2009

### exitwound

Re: Coulomb's Law on a line....again.

$$E=\int_0^{\pi/2} dE sin(\Theta)$$

$$E=\frac{k\lambda}{R}\int_0^{\pi/2}sin\Theta d\Theta$$

$$E=\frac{k\lambda}{R}(-cos\Theta|_{\pi/2}^{\pi})$$ \\\\or from 0-->pi/2

$$E=\frac{k\lambda}{R}(0-1)$$

$$E=\frac{-k\lambda}{R}$$

Which is the same magnitude as the upper half, which makes sense.

(I have to get to bed. I'll check back in the morning. Thanks for the help so far.)

14. Aug 29, 2009

### gabbagabbahey

Re: Coulomb's Law on a line....again.

Right, so the total field is____?

15. Aug 29, 2009

### exitwound

Re: Coulomb's Law on a line....again.

$$E=\frac{-2k\lambda}{R}$$

where $\lambda$ = Q/L where L= R*pi/2

16. Aug 29, 2009

### gabbagabbahey

Re: Coulomb's Law on a line....again.

Right, now plug in the numbers....what do you get for |E| ?

17. Aug 30, 2009

### exitwound

Re: Coulomb's Law on a line....again.

-280.14 n/c

Last edited: Aug 30, 2009