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greenskyy

- 17

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## Homework Statement

Here is my beautiful drawing from mspaint.

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

## Homework Equations

E = ( k dq ) / r^2

[tex]\lambda=dq/ds[/tex]

## 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:

[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 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.

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