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I have the correct answer just the wrong direction

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

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.

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

So

[tex]

E = \frac{k_{e}\cdot \lambda}{r} \int d\theta\cdot\sin\theta}

[/tex]

(integral is from zero to pi)

[tex]

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

[/tex]

[tex]

E = \frac{-2\cdot k_{e}\cdot Q}{L\cdot r} \ \ \mbox{as \lambda = \frac{Q}{L}}

[/tex]

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?

## 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.

## Homework Equations

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.

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

So

[tex]

E = \frac{k_{e}\cdot \lambda}{r} \int d\theta\cdot\sin\theta}

[/tex]

(integral is from zero to pi)

[tex]

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

[/tex]

[tex]

E = \frac{-2\cdot k_{e}\cdot Q}{L\cdot r} \ \ \mbox{as \lambda = \frac{Q}{L}}

[/tex]

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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