# Electric field at center of semi-circular bent rod of charge

I have the correct answer just the wrong direction

## Homework Statement

A uniformly charged insulating rod of length 14.0cm is bent into the shape of a semicircle as shown in figure XXX. The rod has a total charge of 7.5 micro C. Find the magnitude and direction of the electric field at P, the center of the semicircle.

dE = k.dq.1/r^2

## The Attempt at a Solution

I set things up like this
http://img210.imageshack.us/img210/4690/physprobub2.png [Broken]

Since the vertical components would cancel I just looked at the x components.

$$dE_{x} = \frac{k_{e}\cdot dq\cdot\sin\theta}{r^2}$$

$$= \frac{k_{e}\cdot dx\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dq = dx\cdot\lambda}$$

$$= \frac{k_{e}\cdot r\cdot d\theta\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dx = ds = r\cdot d\theta}$$

$$= \frac{k_{e}\cdot d\theta\cdot\lambda\cdot\sin\theta}{r} \ \ \mbox{as dx = ds = r\cdot d\theta}$$

So
$$E = \frac{k_{e}\cdot \lambda}{r} \int d\theta\cdot\sin\theta}$$
(integral is from zero to pi)

$$E = \frac{k_{e}\cdot \lambda}{r} ( -1 - 1)$$

$$E = \frac{-2\cdot k_{e}\cdot Q}{L\cdot r} \ \ \mbox{as \lambda = \frac{Q}{L}}$$
When I substitute the values is I get the correct answer but it's negative. The bent rod is positive so the resultant electric field should be AWAY from the semicircle and therefor positve?!?

What have I done wrong?

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Dick
I have gotten to the point of $$E = \frac{2\cdot k_{e}\cdot Q}{L\cdot r} \$$
but now my question is what is the r value? Of what my reasoning is telling me, it should be $$E = \frac{2\cdot k_{e}\cdot Q}{L^{2}} \$$