# Electric field at the semi-circle's center

## 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?

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.

Related Introductory Physics Homework Help News on Phys.org
TSny
Homework Helper
Gold Member
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.

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
Homework Helper
Gold Member
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.