Solving the Uniform Rod Problem: Calculating Force on a Point Charge

[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\lambdaqcos\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

 

[Doc Al]

_______-^y
________l
________l
________l
________l
________________
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\lambdaqcos\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...

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.