
#1
Jan1911, 09:00 PM

P: 17

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 [tex]\lambda=dq/ds[/tex] 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 southwest. 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: [tex]dE=\frac{k*dq}{R^{2}}[/tex] [tex]=\frac{k\lambda R d\theta}{R^{2}}[/tex] [tex]=\frac{k\lambda d\theta}{R}[/tex] 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 topleft to downright (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: Rereading 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. 



#2
Jan1911, 11:07 PM

HW Helper
PF Gold
P: 1,848

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. 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 [tex] \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 [/tex] [tex] \hat r = \cos \theta \hat x + \sin \theta \hat y [/tex] Integrate over the four different sections separately, then add everything together (vector summation, of course  keep your 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! 



#3
Jan2011, 01:05 PM

P: 17

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 timeconsuming, 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 = [tex]\sqrt{2}[/tex]. 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. 



#4
Jan2011, 06:12 PM

HW Helper
PF Gold
P: 1,848

Calculating the electric field line at a point with a circular field chargeAnd I think you mean E_{y} and E_{x} rather than dE_{y} and dE_{x} 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 [itex] \sqrt{2} [/itex] 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). 



#5
Jan2011, 07:42 PM

P: 17

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. 


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