# Calculating the electric field line at a point with a circular field charge

1. Jan 19, 2011

### greenskyy

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

Here is my beautiful drawing from mspaint.

Using direct integration, find the electric field E at point P due to the line charges.

2. Relevant equations

E = ( k dq ) / r^2

$$\lambda=dq/ds$$

3. The attempt at a solution

All I've been able to do up to this point is calculate the direction that the field charge would be. The +2Q would cancel out the +Q, leaving a +Q at the top, and a -Q on the left. The forces from these two charges would cause the field to point south-west.

I have been at a complete wall when it comes to setting up the integral for these point charge problems. I stumble around arbitrarily plugging in and substituting equations, not really knowing the meaning behind what I'm doing. If I did, I would probably know what to do here. If anyone could offer some enlightenment I would be very grateful.

The work I've done so far is as follows:

$$dE=\frac{k*dq}{R^{2}}$$

$$=\frac{k\lambda R d\theta}{R^{2}}$$

$$=\frac{k\lambda d\theta}{R}$$

I then, completely shooting in the dark, added a cos(theta), and integrated from 135 degrees from 315 degrees. I also concluded that a line of symmetry exists going from top-left to down-right (like y=-x). I cannot see how this can help me in this problem, but I'm sure it does.

Not knowing these problems well wasn't much of an issue when the field charge was one sign, but in this case there is both a positive and negative charge. Again, if anyone could offer some help I would be very grateful. Thanks!!

EDIT:
Re-reading my post, I don't think I'm as clueless as I let on. Sorry, but its just how I feel at the moment =( . I understand that we imagine there being infinitesimally small slices ds taken out of the ring, acting as a point charge, with the integral adding up all the small ds's. We use dtheta to change the displacement of the charge's angle to P. Knowing all this information, I really still can't grasp how to handle these two opposite charges.

Last edited: Jan 19, 2011
2. Jan 19, 2011

### collinsmark

I wouldn't rely on any symmetry for this problem. Let the math do the work for you. Besides, the problem statement instructed that you had to use direct integration. (Meaning if you use properties of symmetry, your instructor might ding you a few points. Just work it out the long way.)

For now, the only thing you really need to get right is the direction of the electric field from the infinitesimal charge slice dQ to the test point, P. This infinitesimal electric field is called dE. But again, don't worry about symmetry! Treat dQ as though its a point charge. (And thus the direction of dE is the direction from dQ to P.
'Looks good to me ...... almost.

The only things missing is to recognize that dE is a vector, and it has direction. First off, let's define the direction of the vector R, as you've defined it in your figure. It has a magnitude of R and a direction of $$\hat r [/itex]. The direction of the electric field is from the charge to the test point. So in this case it's in the opposite direction of [tex] \hat r [/itex]. So modifying your equation every so slightly, [tex] \vec{dE} = -k \frac{\lambda d\theta}{R} \hat r$$
You don't need to take a shot in the dark; you just need to convert the unit vector (direction vector) from polar to Cartesian coordinates.

$$\hat r = \cos \theta \hat x + \sin \theta \hat y$$

Here is where your method needs a lot of work. Break things up into 4 different integrations. (And in a sense this is really 8 total integrations, since you can integrate the x- and y- components separately for each section.)
Integrate over the four different sections separately, then add everything together (vector summation, of course -- keep your x- and y- components separate).
• Integrate from -45o to +45o with q = 0.
• Integrate from 45o to +135o with q = 2Q.
• Integrate from 135o to +225o with q = -Q.
• Integrate from 225o to +315o with q = Q.
Then sum everything together (vector sum -- keeping the x- and y- components separate).

You'll also need to create a relationship between q and λ before you're finished (or before you integrate), so you can express λ in terms of Q (since the relationship is different for each section).

Good luck!

Last edited: Jan 19, 2011
3. Jan 20, 2011

### greenskyy

Collinsmark, your post was everything that I had hoped for, and more. Thank you so much!! It seems the classes I've gone to have focused so much on shortcuts that I never learned the correct way to go about these. Breaking the problem out into all of its components was time-consuming, but made sense. But, let me show some of my new work to see if I'm doing it correctly.

I came up with the equation

( 2 k Q ) / ( pi R^2 )

For the +Q and -Q charges, and 4kQ in the numerator for the +2Q charge. I then wrote out one big equation for dE summing all the different charge components with three integrals. I then separated this integral into two integrals, dEx containing the cosines, and dEy containing the sines.

After everything was said and done, dEy = dEx = $$-\sqrt{2}$$. Giving a magnitude of 2 and a direction of 225 degrees.

The largest difficulty was making sure I was using the right signs, as I had to take into account the direction of the forces and keep track of sin / cos integration.

Last edited: Jan 20, 2011
4. Jan 20, 2011

### collinsmark

Your direction is correct. Both the x- and y- components have the same magnitude which is negative, meaning the direction is 225 degrees.

And I think you mean Ey and Ex rather than dEy and dEx

But I think you left some things out. Perhaps you factored these "constants" out earlier (before the integration -- pulling them out of the integral, maybe) and forgot to multiply them back.

There is a $\sqrt{2}$ involved in the answer (or 2 after taking the magnitude of the resultant vector), yes. But there is more to it than that! the answer is also a function of Q, k and R (and maybe a π and some other constant thrown in too).
This one was kind of a tricky (tedious) one, I agree. But if you keep your unit vectors present, they will tell you the direction. For example, $$\hat x$$ points in the positive x direction. So $$-\hat x$$ points in the negative x direction.

5. Jan 20, 2011

### greenskyy

Ah yes, that's what I meant, sorry :D . I was in a bit of a haste writing my reply, but my final answer did involve those variables - it was the constant I was focused on. Yeah, during my integration I factored out all the variables, which all ended up being multiplied by -sqrt(2).

Actually, in recitation today the TA did cancel out the +Q leaving just the +Q at the top, and proceeded from there. I'm glad I did it the long drawn out way though, because too many shortcuts were being taken by me without knowing what I was skipping.