Question

Anyone familiar with electrostatics? Need some help. Question and my attempt coming up.

Answer 1

A uniform line charge
\[\rho _{l}=\rho _{o}\]
extends from -a to +a along the x-axis. Find the electric field at an arbitrary point P.

My attempt:
Electric Field eqn.
\[E=\frac{ q<r-r \prime> }{ 4\pi \epsilon |r-r'|^{3} }\]

r(vector) = \[x _{xhat}+y _{yhat}\]

r'(vector)=
x' xhat <---- I'm not sure about this.

Answer 2

You would be much better off calculating the electric potential first.

Answer 3

Though if the point P is truly arbitrary, the answer is going to be rather complicated. If it's restricted to the y-axis then it's a different story...

Answer 4

It's arbitrary. I got an insane answer. After that, I have to calculate the potential at an arbitrary point with respect to inf.

Answer 5

I forgot how to find the potential though.

Answer 6

\[ \Phi(\vec r) = \int \frac{\rho}{4\pi \epsilon_0 |r-r'| }dr' \]

Answer 7

If the point was arbitrary, why do I assume that the x-components cancel? If the point lied on the y-axis it would.

Answer 8

I'll try that Jemurray3.

Answer 9

Just so I'm clear
r = <x,y>
r' = <x',0>
right?

Answer 10

yes that's right

Answer 11

Alright, and I'm not sure if I can do this, but.
dr' = d(x' xhat) = xhat dx'.
The potential would only point in the x direction, correct?

Answer 12

The potential doesn't point in any direction, it's a scalar.

Answer 13

Oh yeah forgot about that. So...what would dr' equal?

Answer 14

The integral would look like this
\[ \Phi = \frac{\rho}{4\pi \epsilon_0} \int_{-a}^a \frac{dx'}{\sqrt{r^2 + x'^2 -2x\cdot x'}}\]

Answer 15

I think I see what you're saying.
x^2+y^2 = r^2.
dr' = dx'?

Answer 16

Now, the key to evaluating this integral will be noting that the bottom is
\[ \sqrt{x^2 + y^2 + x'^2 - 2x\cdot x'} = \sqrt{(x-x')^2 + y^2} \]
Then, make the substitution u = x-x' , du = -dx' to yield
\[ \Phi(\vec{r}) = \frac{\rho}{4\pi \epsilon_0 } \int_{x-a}^{x+a} \frac{du}{\sqrt{u^2+y^2}} \]

Answer 17

no, r is just the position vector <x,y>, it has nothing to do with polar coordinates or circles.

Answer 18

You have to be careful with your minus signs. Anyway, moving on from there, you can evaluate this integral via trigonometric substitution. Let
\[ u = y\cdot \tan(z), du = y\cdot\sec^2(z) dz\]
This yields for the integral
\[\int \frac{y\sec^2(z) dz}{y\cdot \sec (z) } = \int \sec(z) dz = \ln|\sec(z) + \tan(z) | \]

Answer 19

now recall that sec(z) = sqrt( 1 + tan^2(z)) and that tan(z) = u/y. Then we have that the antiderivative becomes
\[ \ln|\sqrt{1+u^2/y^2} + u/y| = \ln |(\sqrt{u^2+y^2}+ u)/y| = \ln|\sqrt{u^2+y^2}+u| - \ln|y| \]

Answer 20

the end of that should be - ln(y). Recall that u goes from (x-a) to (x+a) so plugging those limits in we find that the integral is equal to
\[ \ln \left | \frac{\sqrt{(x+a)^2+y^2}+(x+a)}{\sqrt{(x-a)^2+y^2}+(x-a) }\right| \]

Answer 21

Uhm, holy ****. Did you do that by yourself? I just plugged the integral into wolfram and got that answer.

Answer 22

Oh good. I solved it in my head as I was typing it in, thanks for confirming.

Answer 23

I have lots of practice.

Answer 24

Anyway, that's the potential for this horrible mess. You can get the x and y electric field components just by taking the derivatives with respect to x and y respectively, which would be pretty messy, but hey, you gotta do what you gotta do.

Answer 25

Pretty effin impressive. Alright, so the potential will be that times k(rho) and the electric field will be the ...negative derivative? If I'm remembering correctly?

Answer 26

that's correct.

Answer 27

Alright, cool thank you so much.

Answer 28

No problem. I'm going to keep going with this and take the derivative with respect to y and then show you something mildly cool, which can help you check your answers to these problems.

Answer 29

I can just plug them into wolfram if you want me to.

Answer 30

Okay SO
\[ \frac{\partial}{\partial y} \sqrt{(x\pm a) + y^2} +(x\pm a) = \frac{y}{\sqrt{(x\pm a)^2 + y^2}}\]

Answer 31

Yeah, so could I, but it's just to illustrate a point :)

Answer 32

Alright, lol

Answer 33

The partial of y takes a lot of steps and I end up with a very long answer when plugging it in.

Answer 34

You don't HAVE to do it if you don't want. You've helped a lot already.

Answer 35

splitting the logarithm up and taking the derivative of each part (we can save the minus sign for later) gives
\[ \frac{\partial}{\partial y} \log( \sqrt{(x\pm a)^2 + y^2}+(x\pm a)) = \]
\[ \frac{1}{\sqrt{(x\pm a)^2 +y^2} + (x\pm a)} \cdot \frac{y}{\sqrt{(x\pm a)^2 + y^2} }\]

so putting in the minus sign and subtracting gives
\[ \frac{1}{\sqrt{(x- a)^2 +y^2} + (x- a)} \cdot \frac{y}{\sqrt{(x- a)^2 + y^2} }-\]
\[\frac{1}{\sqrt{(x+ a)^2 +y^2} + (x+a)} \cdot \frac{y}{\sqrt{(x+a)^2 + y^2} }\]

That's really gross. However, what happens when a -> infinity?
Taking the limit gives you the answer

\[\vec{E} = \frac{k\rho}{y}\hat y = \frac{\rho}{2\pi\epsilon_0 y} \hat y\]
which you should recognize as the electric field from an infinite line charge.

Answer 36

That's interesting, but is that the answer? Am I supposed to assume that a goes to infinity?

Answer 37

No, it's a cross check. You would obviously want the limit as a->infinity to produce an infinite line charge, and the limit as a-> zero to produce a point charge. When you're doing complicated problems it's a good idea to take quick limits to check things like this.

Answer 38

Alright, that makes sense. I guess that's what they mean when they say 'sanity check'? Thank you for your help, man. I wish I could give you more medals lol

Answer 39

that's exactly what they mean. No problem.