Electric field in the center of an arc

[SetepenSeth]

Homework Statement

Consider a conductor wire with a charge Q uniformly distributed, shaped in the form of an arc of radius R and amplitude 2A (were A is a given number between 0 and π).

Find the value of the electric field in the center of the arc.

Homework Equations

$E(P)= \int K_e (dq/r^2)$

Where $K_e$ is the electrostatic constant[/B]

The Attempt at a Solution

[/B]
Since the charge is uniformly distributed I'm considering the charge density as λ= Q/L thus Q= λL, where L = 2A. The point P is the center of the arc.

So the integral becomes

$E(P)= \int K_e (λL/R^2)$

Getting out the constants, integrating on dL and selecting the limits 0 - L=2A

$E(P)= K_e λ/R^2 \int_0^L LdL$

And solving:

$E(P)= 2A K_e λ / R^2$

However, if I leave it in terms of Q and Pi, the expression oversimplifies and the term 2A goes away

$E(P)= 2A Q / 4πε_0 R^2 2A$ $=Q/4πε_0 R^2$

So I got the serious feeling I'm getting something wrong here. Perhaps integrate in terms of the angle?

Any advise would be appreciated

 

[haruspex]

where L = 2A

What about R?

So the integral becomes
$E(P)= \int K_e (λL/R^2)$

E is a vector. You need to integrate it as such. An integral is a sum. How do you add vectors?

 

[SetepenSeth]

I think I'll start from scratch cause I don't follow

 

[haruspex]

I think I'll start from scratch cause I don't follow

For example, consider the case where the arc goes through 180 degrees. The field at the centre from the charge at one end of the arc is equal and opposite to that from the charge at the other end of the arc, so those cancel.
Choose a coordinate system and write the field as components in that system. You can then integrate each component separately.