# Electric field from a charged rod

Ok, i'm inexperienced with the latex code, so excuse me.

I'm trying to derive the expression for the electric field at a point, along the perpendicular bisector of a charged rod, of known total charge, length L, a perpendicular distance y from the rod.

The horizontal components cancel,
$$\frac{qy}{4\pi \varepsilon_0 L} \int_0^L \frac{dx}{((\frac{L}{2}-x)^2 +y^2)^{3/2}}$$
I don't think this is right, because that seems a nasty integral. Any ideas?

Apologies if you think this is the wrong section. Thanks.

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I finally got the tex perfect. Can somebody please help me as soon as possible? Thanks alot.

krab
The integral's not too bad.. Hint:
$$\int (1+x^2)^{-3/2}dx=x/\sqrt{1+x^2}$$

ehild
Homework Helper
MaximumTaco said:
Ok, i'm inexperienced with the latex code, so excuse me.

I'm trying to derive the expression for the electric field at a point, along the perpendicular bisector of a charged rod, of known total charge, length L, a perpendicular distance y from the rod.

The horizontal components cancel,
$$\frac{qy}{4\pi \varepsilon_0 L} \int_0^L \frac{dx}{((\frac{L}{2}-x)^2 +y^2)^{3/2}}$$
I don't think this is right, because that seems a nasty integral. Any ideas?

You are right, this is not right, and the integral is nasty.
You get a much nicer integral if you place your origin at the middle of the rod, and use the angle $$\varphi$$ instead of x. See pic.

$$r=\frac{y}{\cos(\varphi )} \mbox{ , } x = y \tan(\varphi )$$....

ehild

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ehild
Homework Helper
ehild said:
You are right, this is not right, and the integral is nasty.

ehild

Sorry, the formula was right... But it is still nasty.

ehild