Electric field at the semi-circle's center

[Colts]

Homework Statement

Charge Q is uniformly distributed around the perimeter of a semicircle of radius R. What is the electric field at the circle's center?

Homework Equations

E=kQ/R^2

The Attempt at a Solution

λ=Q/(∏R)

∫(kλ)/R^2 dθ from 0 to pi

and get the answer to be (kQ)/R^3

or is it

∫(kλ)θR/R^2 dθ from 0 to pi

and get (kλ∏^2)/2R

The second one I was trying to put the arc length in, but I'm not sure if you need to or not.

 

[TSny]

Remember that electric field is a vector quantity. Different elements of the semicircle produce electric field at the center in different directions. You need to take that into account. You have probably seen other examples where this is true, such as the field from a straight line of charge.

 

[Colts]

I forgot about that. Thanks. So Since it is a semi-circle. All the charges of the i or j direction (depending on how the circle is orientated) will cancel out. So would I multiply the first integral by cos(θ) to get the "vertical" component (once again depending how it is orientated, but I am thinking of it as a bowl right now). But what would the range of the integral? -pi/2 to pi/2?

 

[TSny]

I forgot about that. Thanks. So Since it is a semi-circle. All the charges of the i or j direction (depending on how the circle is orientated) will cancel out. So would I multiply the first integral by cos(θ) to get the "vertical" component (once again depending how it is orientated, but I am thinking of it as a bowl right now). But what would the range of the integral? -pi/2 to pi/2?

Yes, that sounds good.

 

[Nickie]

I would like to clarify that the electric field at the center of a semicircle of uniformly distributed charge is not dependent on the arc length. The correct answer is (kQ)/R^3, as the electric field at the center is only affected by the total charge Q and the distance R from the center. The second equation you provided may be useful for calculating the electric field at a point on the semicircle's perimeter, but it is not necessary for finding the field at the center. It is important to understand that the electric field is a vector quantity, so it has both magnitude and direction. At the center of the semicircle, the electric field will point directly towards the center of the circle, with a magnitude given by (kQ)/R^3. I hope this clarifies any confusion and helps you in your calculations.