# Electric Field due to Line of Charge

1. Jan 30, 2007

### cd80187

In Fig. 22-41, two curved plastic rods, one of charge +q and the other of charge -q, form a circle of radius R = 3.28 cm in an xy plane. The x axis passes through both of the connecting points, and the charge is distributed uniformly on both rods. If q = 11.8 pC, what is the magnitude of the electric field produced at P, the center of the circle?

The Figure is simply an x and y axis, with a circle being drawn on it, the top half of the circle being + and the Lower half being -.

The equations that I am given is E = (q x z)/ (4pi x permitivity constant)(z squared + R squared) ^ 3/2 where q is the charge, z is the vertical component height point P is above the circle, and R is the radius of the circle.

For this problem, I am unsure how to use this equation. They are asking me to find the charge in the middle of the circle, therefore there is no vertical component (z), which would give me an answer of 0, which is wrong. So I am unsure of how to go about solving this, or if there is another equation that I am missing or something.

2. Jan 30, 2007

### Tom Mattson

Staff Emeritus
OK.

Can I ask why you think that this equation goes with your problem? If you're in the xy plane, then z=0. That would mean that the electric field is 0 at the center, which is clearly wrong.

Rather than rely on a equation, let's try to analyze this using basic principles. Consider first the differential electric field due to the top half (call it $d\vec{E}_+$), and then the differential electric field due to the bottom half (call it $d\vec{E}_-$). Then integrate them and take their vector sum.

Start by looking at the top half (the one with positive charge). Consider a differential element of charge $dq=\lambda ds$, where $ds$ is a differential arc length.

By symmetry, which component of $d\vec{E}_+$ cancel out?
Can you then write down an expression for $d\vec{E}_+$?
Can you then integrate that expression?
Can you then move on to $d\vec{E}_-$?

Try to answer those questions, and let us know where you get stuck.

3. Jan 30, 2007

### cd80187

OK, i'm not sure about this, but shouldn't everything cancel out, except for the one line that is perpindicular to the x axis? (aka the y-axis line).
So then could I just use a point charge equation of (1/ 4pi x permitivity constant) x (q/r squared). But I have no clue how to integrate it, or if it is right for that matter, but I believe that I would simply repeat the process for the negative side.

4. Jan 31, 2007

### Tom Mattson

Staff Emeritus
No, it is the component parallel to the x-axis that will cancel out upon integration. To see this, consider the field due to a point on the positively charged semicircle (not on the y-axis). Then look at the field due to its mirror image in the y-axis. Draw the field vectors due to both points on the semicircle, and you'll see that the horizontal components are equal and opposite, while the vertical components reinforce each other.

OK, that's start. You need to consider a differential segment $ds$ of the arc length of the circle. Let's begin with the positively charged half. Since it has uniform linear charge density $\lambda$, the differential charge $dq$ on that segment is $dq=\lambda ds$. So the differential electric field $d\vec{E}_+$ due to just that segment is:

$$d\vec{E}_+=\frac{1}{4\pi\epsilon_0}\frac{dq}{R^2}[\cos(\theta)\hat{i}+\sin(\theta)\hat{j}]=\frac{1}{4\pi\epsilon_0}\frac{\lambda ds}{R^2}[\cos(\theta)\hat{i}+\sin(\theta)\hat{j}]$$

Now, as has been pointed out already, the $\hat{i}$-component of the differential fields will cancel out when you integrate them. That means that you only need to consider the other part. So you need to get $\sin(\theta)$ in terms of some convenient coordinates. Can you do that?

And once you have found the field for the positive half, you shouldn't need to repeat your reasoning for the negative half. You should be able to find it by symmetry.

Give that a try, and let us know if you get stuck.