Electric field of a semi circle

[madah12]

Homework Statement

[PLAIN]http://img101.imageshack.us/img101/2786/21417885.png
[96]

Homework Equations

The Attempt at a Solution

E= kdq/r^2
dq=Q/(pi a) dx
Ex = 0 , Ey= E sintheta
sin theta = sqrt(a^2 - x^2)/a by Pythagorean theorem
Ey = kQ/(pi a* a^3) integral from -a to a sqrt(a^2 - x^2)dx
= kQ/(pi a^4) *pi a^2/2= kQ/2a^2 but in the manual it says E= 2kQ/pi*a^2

 

[FLms]

Hi,

You could try an integral in function of [tex]\theta[/tex].

 

[madah12]

but I already did it as a function of x why was that wrong?
I think that is ok because every point in the semi circle can be expressed by the x cooridnate

 

[madah12]

any help?

 

[ehild]

dq=Q/(pi a) dx
Ex = 0 , Ey= E sintheta
sin theta = sqrt(a^2 - x^2)/a by Pythagorean theorem
Ey = kQ/(pi a* a^3) integral from -a to a sqrt(a^2 - x^2)dx
= kQ/(pi a^4) *pi a^2/2= kQ/2a^2 but in the manual it says E= 2kQ/pi*a^2

Your dq means uniform charge distribution along x; but it is uniform along the arc. dq=Q/(pi a) (a dtheta). Integrate wit respect to theta. ehild

 

[madah12]

Ey = Q/(pi a * a^2) integral from 0 to pi a*sin theta d theta
= -Q/(pi a^2) (-1-1) = 2Q/(pi*a^2)

 

[ehild]

Do not forget k.

ehild

 

[madah12]

oh ok but so if we are not integrating along the x-axis then are integrating along the circle?

but how can we know that dq= Q/(pi a) (a dtheta) represents integration over the arc? I did it because you said that it did but I don't understand it.

 

[madah12]

what I am asking is that how can I know what dq is if I am not integrating over a straight line
I mean I was never taught how to integrate over a curved line.

 

[FLms]

You are integrating over the semi-circle because you are varying the angle. Every infinitesimal charge dQ will have a different angle in relation to the point P, so you have to count it.

So:

[tex]dQ = \lambda dl [/tex]
[tex]dl = r d \theta[/tex]

[tex]\lambda = \frac{Q}{l}[/tex]

(lambda is the linear density.)

 

[ehild]

The problem said that the charge is uniformly distributed along the semicircle. It means that a tiny piece of the arc of length dl has the charge dq=Q/(pi a) dl. But the length of arc is proportional to the angle in radians: dl = radius times angle,
dl = a*d(theta).
You know that the integration variable can be changed. It is easier to integrate along the length of arc than along x.

All the dq charges along the semicircle contribute to the electric field at the centre, and you have to sum up their contributions.
All these little charges belong to a certain angle theta and their electric field is parallel to the radius. The x components cancel and the y components are kdq/a^2 sin(theta). Replace dq by Q/(pi a)* dl, dl by a*dtheta, and instead of the sum, you do integration.

ehild