# Uniform rod problem

1. Sep 6, 2007

### venom_h

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l________________l--------> x

Sorry for the bad picture, but suppose there's a uniform rod of charge has a length L and a net charge +Q, find the force that this rod exert on a point charge q placed at (0,y).

Ok, I found by symmetry that the F(x) is 0.
$$\lambda$$=dQ/dl

So, F(y) = $$\int$$ k$$\lambda$$qcos$$\Theta$$(dl)/ (l^2 + y^2) , [-L/2, L/2]

And since y/r=cos$$\Theta$$, r= ysec$$\Theta$$, where r = (l^2 + y^2)

Then it boils down to 2kqy$$\lambda$$$$\int$$ (dl)/ (y^3(sec$$\Theta$$)^3) , [0, L/2]

Then I don't know how to carry on without using the integral table....
But it should be something like this and i don't know why:
F(y) = 2kqy$$\lambda$$$$\int$$(y(sec$$\Theta$$)^2d$$\Theta$$)/ (y^3(sec$$\Theta$$)^3)
and gives 2kqy$$\lambda$$$$\int$$(cos$$\Theta$$) d$$\Theta$$
............. ans= kqQ/ y$$\sqrt{y^2 + (L/2)^2}$$

Anyways, my question is, without using the integral table, how do people actually move on to that next step?

Thanks

Last edited: Sep 7, 2007
2. Sep 7, 2007

### Staff: Mentor

Get rid of those angles! You should have an expression for dF in terms of x & y (I use x as the coordinate of the line charge element):
$$dF = \frac{kq\lambda}{(y^2 + x^2)}\cos\theta \; dx = \frac{kq \lambda x}{(y^2 + x^2)^{3/2}} \; dx$$

You should have no problem integrating this without a table.