How Is the Electric Field Calculated at the Center of a Semicircle of Charge?

[espen180]

Homework Statement

Positive charge Q is uniformly distributed around a semicircle of radius a (See fig). Find the electric field (magnitude and direction) at the center of curvature P.

http://img132.imageshack.us/img132/8278/electricfield1.png

Homework Equations

Electric field of point charge:
$$\vec{E}=k\frac{Q}{r^2}$$

The Attempt at a Solution

Since I suspect the linear charge distribution $$\lambda=\frac{Q}{\pi a}$$ will be troublesome to wirk with here, I will define the angular charge distribution $$\alpha=\frac{Q}{\pi}$$, divide the semicircle into infinitisimal pieces, calculate the electric field for each one and integrate.

Since each piece is a distance $$a$$ away from point $$P$$, the electric field caused by each piece is $$dE=k\frac{dQ}{a^2}$$. $$dQ=\alpha d\theta=\frac{Q d\theta}{\pi}$$, so $$dE=k\frac{Qd\theta}{a^2\pi}$$.

By symmetry, the total electric field in the x-direction is zero, so I only have to find the field in the y-direction, which is given by $$dE_y=dE\cdot \sin \theta=\frac{kQ}{a^2\pi}\sin\theta d\theta$$.

Integrating from $$\theta=0$$ to $$\theta=\pi$$, the result is
$$E_y=\frac{kQ}{a^2\pi}\int_0^\pi \sin\theta d\theta=\frac{kQ}{a^2\pi}\left[-\cos\theta\right]_0^\pi=\frac{2kQ}{a^2\pi}$$.

Since Q is positive, the field is in the negative y-direction.

$$\vec{E_P}=-\frac{2kQ}{a^2\pi}\hat{j}$$

 

[diazona]

What you've done looks perfect to me... any particular reason for asking about it?

 

[espen180]

I've done this sort of problem a number of times with straight lines, and the book did not have an answer key for this paticular problem. I was just wondering if I had the correct answer.