Electric Field Due To a Rectangular Charged Wire

[nikkou]

Hi guys, here's a question i thought i knew how to solve but somehow it doesn't work. been trying and trying for almost a week :-(
here it is:

"A thin wire is bent in the form of a rectangle with length l = 12 cm and width w = 10 cm. The wire is charged up to a uniform charge density of 5.5×10-6 C/m. Find the magnitude of the electric field at a distance of 65 cm from the center of the rectangle along the axis."

http://img73.imageshack.us/img73/2673/11542as4.gif [Broken]

I've tried calculating the sum integral of the field at that point using gauss' law for each side of the frame but i don't get it right. I know how to integrate, so i don't think it's a calculation error, maybe my integral is wrong. or maybe there's a better way?
Thanks!

 

[armis]

What about taking the L and W frames and integrating by angle of each little element ( dL or dW respectively ) to the d using Coulomb's law. You'll get the magnitude of the electric fields created by L and W frames then all what's left is to multiply both by 2 as you have 4 frames and sum up. Wouldn't that work? Although calculations might get tricky

There might be simpler ways to do it though

 

[nikkou]

What about taking the L and W frames and integrating by angle of each little element ( dL or dW respectively ) to the d using Coulomb's law. You'll get the magnitude of the electric fields created by L and W frames then all what's left is to multiply both by 2 as you have 4 frames and sum up. Wouldn't that work? Although calculations might get tricky

There might be simpler ways to do it though

How would i use the angle? i integrated using the distance which i got by the pythagorean theorem.

 

[Nick89]

Don't use Gauss' Law here, there is no appropriate symmetry.

Instead, use Coulomb's law.

Do you know how to use Coulomb's law to calculate the electric field at some distance from a wire?
In this case, you have four wires (two of which are equal so actually only two). The magnitude of the electric field of the four wires combined is ofcourse just the sum of all wires.
The direction can be a bit more tricky, but in this case, since the point A is directly above the center, it will only be upwards.


EDIT
I worked it out quickly and I got an answer of
$$E_{tot} = 51231.5 \, \frac{\text{N}}{\text{C}}$$

Do you have any answers, is this correct?

If it is I can tell you how I did it but it would be better if you figured it out yourself! (I used the method I described above)

If you don't know how to calculate the field of a wire here is what I believe is the answer, but if you don't know how to do this this question is impossible I think...

$$\vec{E} = \frac{1}{2 \pi \epsilon_0} \frac{\lambda a}{x \sqrt{ x^2 + a^2}} \hat{i}$$
Here, $\lambda$ is the charge distribution, $x$ is the distance from the wire (0.65m), $a$ is half of the length of the wire (0.12/2 = 0.06 m (or 0.05 for W)) and $\hat{i}$ is the unit vector in the x-direction.
(Note, this is the fieldof ONE straight wire, not the total field required by this question!)

 

[armis]

Then what's the problem?

I think your integral should more or less look like this:
$$2\int^6_0\frac{\rho{dr}}{4\pi\epsilon_0(d^2+r^2)}$$

Where $$\rho$$ is the charge density while dr is a small element of L. Looks allright, is that the way you did it?

 

[Nick89]

There is no problem? I was simply trying to get nikkou started by giving him some hints about what he needed to do.

That's more or less what I did, not exactly. I used:

$$E_x = \frac{1}{4 \pi \epsilon_0} \frac{Qx}{2a} \int_{-a}^a \frac{dy}{(x^2 + y^2)^{3/2}} = \frac{Q}{4 \pi \epsilon_0} \frac{1}{x \sqrt{x^2 + a^2}}$$
where Q is the total charge: $$Q = 2a \lambda$$.

 

[armis]

ah, sorry Nick I didn't notice your post. That's why I asked what is the problem

 

[armis]

I don't really get your integration process :(
Why do you take the whole charge? Only small parts of the whole charge contribute to the electric field at a given distance. Or am I missing something?

 

[nikkou]

There's something i don't get - the distance you're both using isn't the right one. it's true just for the middle of one of the sides - the distance from [dl] to the point of measurement is changing for any [dl].
or am i totally wrong?

 

[Nick89]

Don't look at the total picture just yet, first try to understand the electric field of a single wire of length 2a. If you understand that, you can apply it to the two seperate wires here and sum the result (but make sure to only count the component in the up-direction while summing!)


For the electric field of a wire here is what I did, I will try to explain as much as I can:

Consider the following image:

The line (yellow) has length 2a and has a positive total charge Q, which is distributed uniformly across the wire.

Coulomb's law allows you to look at a tiny piece of the wire first and then integrate over the whole wire.

Consider the red part of the wire with length dy and charge dQ. If the linear charge density is $\lambda$ then the charge dQ can be expressed as:
$$dQ = \lambda dy = \frac{Q dy}{2a}$$
because $\lambda = \frac{Q}{2a}$.

The distance from the point at dQ to point P is r, which can be expressed as:
$$r^2 = x^2 + y^2$$

Now according to Coulomb's law, the magnitude of the electric field at P due to the little wire segment at dQ is:
$$dE = \frac{1}{4 \pi \epsilon_0} \frac{dQ}{r^2} = \frac{Q}{4 \pi \epsilon_0} \frac{dy}{2a (x^2 + y^2)}$$

Note closely that this field has both x- and y-components!
If you look at the symmetry of the problem it is not hard to see that, eventually, once you take into account the whole wire, the y-components will all cancel each other out. (The top part of the wire will cancel out the y-component of the lower part, etc). Only the x-component will be present. Therefore we can now skip the y-component entirely and only concentrate on the x-component.

The x-component of dE is given in terms of the angle $\theta$ as:
$$dE_x = dE \cos \theta = dE \frac{x}{\sqrt{x^2+y^2}}$$

Substituting dE we found before we get:
$$dE_x = \frac{Q}{4 \pi \epsilon_0} \frac{x \, dy}{2a (x^2 + y^2)^{3/2}}$$

Integrating this over the entire wire (y from -a to a) we get the following integral:
$$E_x = \frac{1}{4 \pi \epsilon_0} \frac{Qx}2a} \int_{-a}^a \frac{dy}{(x^2 + y^2)^{3/2}}$$

This is a standard integral, look it up in an integration table if you don't know it.

Calculating the integral we finally get:
$$E_x = \frac{1}{4 \pi \epsilon_0} \frac{Q}{x \sqrt{x^2 + a^2}}$$

So this is the electric field at a radial distance x from the center of a wire with length 2a and uniform charge Q.
Now what stops you from using this equation exactly to calculate the electric field at the point A in your image? Surely you can calculate the total charge Q and you know the distance x for both wires.

What seems to be bothering you is that the point A is not directly above the wire. This does not matter! The wire is completely symmetrical about it's axis so you can rotate it about it's axis without anything changing. Just make sure to use the right distances in each calculation and you will be fine!

Try calculating the electric field at point A due to one of the 'L' wires (length 2a = 0.12m). I got an electric field of:
E = 13901.6 N/C
(Note, this is not only the up-component yet)
See if you can get the same.

You only have to sum them up afterwards, although there is one catch.

Just like I used the cosine of the angle in the above derivation, you also have to take into account the fact that the field of one of the wires in point A is not directed upward but rather a little to the side. You only need to sum the upward component of the field since due to the symmetry, all the 'to the side' components will cancel out. (This is only a tiny change in the answer though, but that's only because the angle is so small!)

 

[armis]

Wow, nicely done Nick!
That was the problem with my integral. I didn't take into acount the symmetry of the y component.
By the way what program did you use to draw the image? I imagine it was not paint

 

[Nick89]

Did you get the right answer now?

No, I didn't use Paint to draw the image :P I use Photoshop.

 

[armis]

I was not the one asking for help, although I could certainly use some :)
nikkou started this thread. But I have no doubt he will get it right after your excellent explanation

Nice, I should use one too. Allright, there is that photoshop I have got...

 

[Nick89]

Oh right sorry ^^.

I use Photoshop CS3 which is the latest version. My dad got it at work so I was in luck; if you want to buy it legally it is really expensive... But there are free versions (GIMP for example) which can do most of the basics and are more than enough to make images like this...

 

[armis]

Thanks. But I do have CS2 somehwere around here... I'll get GIMP though, looks quite nice

 

[nikkou]

WOW! That's amazing nick, thanks so much!
you definitely got what it takes to explain this stuff, it worked perfectly!
the only thing i did different is that i expressed the distance as
$$\sqrt{ (\frac{l}{2})^{2}+(\frac{w}{2})^{2}+d^{2} }$$
which is the "real" distance from the wire to the point of interest.
then i integrated once with l as the variable and once with w.
but i only managed to solve it after reading your post, i read it and then tried to reconstruct it alone and it worked like a charm.

So thanks very much again and keep up the good work, this place is awesome!

 

[Nick89]

Did you use that distance in your integration or afterwards while summing the fields of each independend wire?

Can you state the final answer of this problem, so I can see if my method worked the way I had it in mind?


EDIT
I worked it out and got a final answer of:
E = 50751.67 N/C

This should be the field at point A due to all four wires. Is this correct?

 

[nikkou]

Ok, after reading your post and starting over with the intention to recreate what you wrote there, here's what i ended up with:

as i said the distance i used in the integration is:
$$\sqrt{(\frac{l}{2})^{2}+(\frac{w}{2})^{2}+d^{2}}$$
(this is the distance from dl/dw to p)

and my integral was:

$$E=k\lambda d*\int((\frac{1}{(\frac{l}{2})^{2}+(\frac{w}{2})^{2}+d^{2}})* cos\alpha) dl$$

where
$$cos\alpha = \frac{d}{\sqrt{(\frac{l}{2})^{2}+(\frac{w}{2})^{2}+d^{2}}}$$

and after integration:
$$E=k\lambda d*\left|\frac{\frac{l}{2}}{(d^{2}+(\frac{w}{2})^{2})*\sqrt{(\frac{l}{2})^{2}+(\frac{w}{2})^{2}+d^{2}}}\right|^{l}_{-l}$$

i did the same for the w side, multiplied both by 2 and the final answer i got was 50904.9 N/C. the given solution is 5*10^4 so yipee to both of us!
Thanks again for your help! appreciate it a lot.

 

[Gear300]

Considering that the lengths of the wire are relatively finite, Gauss's Law may not be the best choice. Using a summation of increments of charge dq could probably help out better. To take into account vectors in the integral, you could note that there is no net electric field horizontally (they symmetrically cancel out); so the net electric field is a result of the field vertically.

 

[Nick89]

I'm not sure if your answer is entirely correct, I think you forgot to only sum the up-components of the E-fields of the four wires. If you do this you will get an extra:
$$\cos \left( \arctan \left( \frac{5}{65} \right) \right)$$ (and 6/65 for the other side) which is about 0.996, giving you a slightly smaller answer.

If all they do is give you the answer 5*10^4 N/C then imo that's really bad... You can make lots of little mistakes and still get 5*10^4 rounded... They should really give you the approach in the solution aswell as the final answer...

 

[nikkou]

I'm not sure if your answer is entirely correct, I think you forgot to only sum the up-components of the E-fields of the four wires. If you do this you will get an extra:
$$\cos \left( \arctan \left( \frac{5}{65} \right) \right)$$ (and 6/65 for the other side) which is about 0.996, giving you a slightly smaller answer.

If all they do is give you the answer 5*10^4 N/C then imo that's really bad... You can make lots of little mistakes and still get 5*10^4 rounded... They should really give you the approach in the solution aswell as the final answer...

but my integration is done only on the vertical components!
$$E=k\lambda d*\int((\frac{1}{(\frac{l}{2})^{2}+(\frac{w}{2})^{ 2}+d^{2}})* cos\alpha) dl$$
the $$cos\alpha$$ takes care of that.

I don't have the exact official answer to the the numbers given on my original post. i do have it for these:
l = 35 cm
w = 34 cm
$$\lambda$$=7.000×10-6 C/m
distance of 80 cm

the answer is 1.24×10^5 V/m which is what i got using my way.

 

[Nick89]

Ah yes I see now. I got 1.24 aswell on the second example so I guess were right :)