Electric field of a continuously charged semicircle rod at center

[JosephK]

Homework Statement

A uniformly charged insulating rod of length 14.0 cm is bent into the shape of a semicircle. The rod has a total charge of -7.50[itex]\mu[/itex]C. Find (a) the magnitude and (b) the direction of the electric field at O, the center of the semicircle.

Homework Equations

d[itex]\overline{E}[/itex] = [itex]\frac{k dq}{r^2}[/itex][itex]\widehat{r}[/itex]

The Attempt at a Solution

This is a continuous charge distribution. The electric field points in the +x direction.

I made dq = [itex]\lambda[/itex]d
and integrated
[itex]\overline{E}[/itex] = [itex]\int[/itex][itex]\frac{k dq}{r^2}[/itex][itex]\widehat{r}[/itex]

from 0 to l, where l is the length of the rod

This is incorrect because r is constant.
R can be found through geometry.

I did not understand how to use cos[itex]\theta[/itex].
I reasoned at the ends of the rod, the angle is 90.
I did not figure out what angle it is.

 

[danielakkerma]

Hi, if you don't mind, I made the effort to sketch your problem.
Now, if you take a look at the diagram, a small element of the field [itex] d\vec{E} [/itex] is created by a minor element dl.
Since the rod's initial length was l, that remains the entire length of the arc. The radius of the circle is now, by l = pi*R, R = l/Pi;
A small element, a fraction of the semicircle is now, dl = R*d(theta);
Assuming a linear charge distribution: [itex] \lambda = \frac{Q}{l} [/itex], We arrive at the
following:
[itex] d\vec{E} = \frac{k}{r^2}dq [/itex], R is constant, as you said.
But there are components to the E field, on the x, y axes respectively, axes which can be chosed arbitrarily, but must be maintained.
Let's take, per convention, The horizontal as x, vertical as y(note, direction donward taken as positive!).
We then get dE_x = dE*cos(theta), dE_y = dE*Sin(theta).
Therefore:
[itex]
\Large
E_x = \displaystyle \int_0^E{dE\cos(\theta)} = \int_0^\pi{\frac{k\cos(\theta)}{R^2}\lambda R d\theta} = 0;
[/itex]
[itex]
\Large
E_y = \displaystyle \int_0^E{dE\sin(\theta)} = \int_0^\pi{\frac{k\sin(\theta)}{R^2}\lambda R d\theta} = 2\frac{k}{R}\lambda
[/itex]
Plug in R, and lambda above, and you're done!
Daniel