# Electric field of an infinitely long charge carrying wire

Tags:
1. Aug 8, 2017

### Rahulrj

1. The problem statement, all variables and given/known data
Find the electric field of an infinitely long straight wire of charge $\lambda$ C/m at a point $r= ix+jy$

2. Relevant equations
$\int E.da = \frac {Q}{\epsilon_0}$
$E= \frac{\int dq}{4\pi\epsilon_0 r^2}$r

3. The attempt at a solution
Drawing a cylindrical gaussian surface of radius 's' I can find E, using the gauss's law:
$E.2 \pi s L = \frac {\lambda L}{\epsilon_0}$
$E = \frac{\lambda}{2\pi s\epsilon_0}$
So is my answer correct if I just substitute the given 'r' into the above equation for 's'? which is $E = \frac{\lambda}{ 2\pi (ix+iy) \epsilon_0}$r (where r is unit vector) I find it bit odd.
Alternatively when I try to find E using the electric field formula from Coloumb's law, I am stuck with the integration as the wire extends from negative to positive infinity and I am not sure of how to include vector into it.
$E= \frac{\int dq}{4\pi\epsilon_0 r^2}$
$dq = \lambda dl$
If i assume wire to be lying on the x - axis, how should I write the separation vector r in the above equation?
Is it $r_* = xi+yj-xi = yj$ thereby $r=y^2$ or just directly substituting $r =xi +yj$ I am just confused with the vectors here.

Last edited: Aug 8, 2017
2. Aug 8, 2017

### Vrbic

Your result coming from Gauss law is quite right. Maybe should be nicer (ix+jy) write as |r| if your r is unit vector.

Here I can advise you to change variable. You can find only one variable describing intensity at r.

3. Aug 8, 2017

### Rahulrj

I did not get what you meant by changing variable.
$dl=dx$
I am just not sure what $r^2$ here is and since integrating limits involve infinity, I am not certain on how to approach this problem in that form.
I follow griffith's electrodynamics and in the example 2 of chapter 2 (mentioned so that you can refer), there is a similar problem its just not infinite line charge, there he solves the problem using vectors, I am not able to grasp that concept well. In the book 'r' is the separation vector given by $r = r_p - r'$ (r' - source position, r - field point) and solution follows by multiplying the unit vector and then finally integrating. However going on the similar line I get separation vector to be $yj$ and problem comes up with the limits.

4. Aug 8, 2017

### Vrbic

I'm not sure if I understand right what is your problem but I mean it this:
1) I don't know if it is typo but this
is wrong (sorry I missed it first). Integration HAS to be over the whole fraction. dq has inside r (or opposite). You are mixing position of point and distance wire - point.

2) Integration over vectors is very dificult. In practice you make an effort to part it to diferent components of the vector. Find a symmetry of the problem (which has to be there if you are able to aplly Gauss law) and integrate it separatly (or find out that some components do not contribute). As in this case.
Hint:
Place your point ix+jy on y-axis (i.e. coordinate (0,R), I change it R is point - constant and r is distance - variable ). You can do it, it is infinit wire, "in the middle" is everywhere, i.e. result is same for all points with y-coordinate R. And then sketch vector of intenzity $dE_1$ in your point R contribuing from arbitrary point "b" (b,0) on a wire (x-axis) and then intezity vecotr $dE_2$ from the point "-b" (-b,0). What can you say about these vectors?

5. Aug 8, 2017

### Rahulrj

Oh sorry that was a typo, I didn't intend it to be like that.
I can right away say that the x components cancel and y components add up.
So now I have the point $r=ix+jy$ at a distance R (perpendicular length) from the line charge at x axis. I can draw a hypotenuse to an element $dx$ on the line charge and write the following:
$dE = \int \frac{\lambda dx}{4\pi \epsilon_0 (x^2+R^2)}$ and multiply with $\cos\theta = R/r$ for y component.So I will get $dE = \int \frac{R\lambda dx}{4\pi \epsilon_0 (x^2+R^2)^\frac{3}{2}}$. Evaluating it for the limits, $E =[\frac{x\lambda}{4\pi \epsilon_0 R(x^2+R^2)^\frac{1}{2}}]_\infty^{-\infty}$ This is where I go wrong and this method is the same way of working out a problem without vectors which is where I think I am making mistake.

6. Aug 8, 2017

### Vrbic

Here is small bug. If you write differential on one side, integral sign is extra.

It is right, but not so easy to interpret. You have to use limit for determining this. Can you find limits $\lim_{x->\infty,-\infty}\frac{x}{\sqrt{x^2+R^2}}$?

But still, if you use your $\theta$ as a variable you integrate from $\pi/4$ to $-\pi/4$. On the other hand you have to be careful in interpretation dq, how find relation dq vs d$\theta$.

7. Aug 8, 2017

### Rahulrj

So if this is right, don't we have to include the information of the point? up till here $r=xi+yj$ is not included in the integration,if I am correct.
Also I am not really sure about using limits here, could you please hint me on that? and I did not understand the choice of $\pi/4$ too.

8. Aug 8, 2017

### Vrbic

Point R=(x,y) in coordinates i,j is constant point (in our case (0,R) ). You choose it at start. If you like it you can substitute it back at the end. As you said, important is only perpendicular distance R so result for point (b,R) is same as for (0,R) (only for infinite wire).

Hint to limit: Square the limit. If you have square of this term are you able to solve it?

What exactly do you not understand? The limits of integration or changing of variable to $\theta$?

9. Aug 8, 2017

### Rahulrj

Okay that makes sense. So to make sure, if the point were $r = 2xi +3yj$, I could use (0,3R) or if I take $dy$, I could use (3R,0) right?
so in this case, the denominator becomes $(x^2+9R^2)$?
In essence, we have to keep one of the coordinate as a constant, am i right?

I can write it as $E =[\frac{2x\lambda}{4\pi \epsilon_0 R(x^2+R^2)^\frac{1}{2}}]_0^{\infty}$
taking limit, $E =[\lim_{c\rightarrow +\infty} \frac{2x\lambda}{4\pi \epsilon_0 R(x^2+R^2)^\frac{1}{2}}]_0^c$
$E =[\lim_{c\rightarrow +\infty} \frac{2\lambda}{4\pi \epsilon_0 R(1+\frac {R^2}{x^2})^\frac{1}{2}}]_0^c$
$E = \frac{2\lambda}{4\pi \epsilon_0 R}$
Hence here R is $ix+iy$ correct?

I did not understand the limits, the reason why you specified $\pi/4$.
If I am right the change of variable is using $\cos\theta = R/r$ right?
but $x^2+R^2$ does not disappear.

Last edited: Aug 8, 2017
10. Aug 8, 2017

### SammyS

Staff Emeritus
What is the location of the wire? Along the x-axis? Along the y-axis? Maybe along he z-axis ?

11. Aug 8, 2017

### Vrbic

(0,3R) is right. Problem is (mayby my) same mark. r as point where you look for field and r distance between point and wire. Let's start to call a point where you look for electric field is $A=(x,y)$. Or as you wrote $A=(2,3)$!! (x,y) is just generall name in (i,j) coordinates. I don't know how to say your x=2x. Better than write (2x,3y) is write (2,3) as a substitution of (x,y). And you choose it at start and it is constant through all calculation.

Interesting, but correct I mean.

And $x^2+R^2$ what is it... just express it from $\cos\theta = R/r$. Just be careful on $dq$ because of you need to express is in $d\theta$

12. Aug 8, 2017

### Rahulrj

I took it be along x-axis.

13. Aug 8, 2017

### Rahulrj

Maybe that's where I am getting confused, the 'r' in the denominator represents the distance between point and wire.
so when a point is given where field is to be calculated say $r_*$, I should just use that point as the 'r' in the denominator correct?

14. Aug 9, 2017

### haruspex

The question surely means along the z axis. If it is a uniform charge along the x axis then the answer will not depend on x, so no point in specifying the point as $\hat ix+\hat jy$.

15. Aug 9, 2017

### Rahulrj

Now that's confusing, In the above posts, I solved it keeping the line charge on x axis and the result I got depended on the y axis which I took to be R.
In the first post, using the flux rule I got $E = \frac{\lambda}{2\pi \epsilon_0 s}$ and I took that s to be the given point which is $\hat ix+\hat jy$, is it not right?
Much of my confusion is because of the vector involved, I don't get the difference between using the given point as 'r' in the denominator and 'r' as a separation vector as used in the text I follow.
So How would the separation vector vary if the line charge is on z axis? would it be $ix+iy$ or something else?
PS: the actual question is to find the total EF at the point \hat ix+\hat jy of two infinite line charge on x and y axis. My intuition is total EF is vector sum of $E_x$ and $E_y$.

16. Aug 9, 2017

### Vrbic

It is right, but as I wrote earlier, it is good look for simplifying. If you realize that important is only distance of point from wire (my R), you can reduce this problem to 1D problem i.e. you can describe whole problem only by one variable. At the end you can go back to the 3D, put wire on z-axis and initial point in x-y plane i.e. write R in terms x and y and unit vector pointing in z-direction.

Am I right haruspex?

17. Aug 9, 2017

### haruspex

Ah, that changes things. Complete information always helps.

18. Aug 9, 2017

### Rahulrj

I don't know if I am confusing myself, according to the solution of the post #9, keeping line on the x axis, the denominator has to be just $y \hat j$ right? not $x\hat i+y \hat j$ because R there was a constant in the y coordinate. so from flux rule, the denominator 's' has to be $y \hat j$ instead of $x\hat i+y \hat j$ as what haruspex said in the post #14.
Therefore according to my understanding, if I put wire on z axis, the denominator would be in terms of x and y agreeing with your conclusion but how come it points in z-direction? the line is in z axis so direction has to be perpendicular to z right?

19. Aug 9, 2017

### haruspex

You are confusing yourself by using x to mean two different things. On the one hand it is a coordinate of a fixed off-axis point, on the other you integrated wrt x along the x axis.
By symmetry, the x coordinate of the fixed point is irrelevant for the field from a uniform charge along the x axis, so you could have specified the point as (0, y). What does that make R equal to?

20. Aug 9, 2017

### Rahulrj

Yes by symmetry x components cancel at the point, so are you agreeing with me that the R should be $y \hat j$?