2-Dimensional Line charge, around a ring.

[MuckThatGuy]

Homework Statement

This is a fairly easy problem, conceptually... but I can't seem to put the numbers together correctly.

A circle with a radius "r" has a linear charge along the circumference with a charge density of $$\lambda=\lambda_{0}\sin(\theta)$$, where $$\lambda_{0}$$ is the max charge along the circumference.

Find:
a.) The direction of the electric field at the center of the ring.
b.) The magnitude.

(They're obviously not looking for a numerical answer)

2. The attempt at a solution

a.) The direction is definitely down, however, I can't seem to 'show' this outside of just my own intuition.

My best attempt at getting everything together is as follows:

$$d\vec{E_{r}}=d\vec{E_{x}}+d\vec{E_{y}}$$

$$d\vec{E_{x}}=k\int\frac{dQ}{r^{2}}=k\int\frac{\lambda \;dL}{r^{2}}=k\int\frac{\lambda_{0}sin(\theta)\;dL}{r^{2}}$$

$$d\vec{E_{y}}=k\int\frac{dQ}{r^{2}}=k\int\frac{\lambda \;dL}{r^{2}}=k\int\frac{\lambda_{0}sin(\theta)\;dL}{r^{2}}$$

When attempting to get dL into terms of $$d\theta$$ is when problems arise. I know it has to be simple, but every method I've tried always ends up with a charge of 0 along both the x and y axis's (which is obviouslly wrong).

ETA: I'm not sure if it just suffices to change dL to the x and y components $$r\;cos(\theta)$$ and $$r\;sin(\theta)$$ respectively. It doesn't feel right to me, but it's the best I can come up with right now. (The book doesn't cover a problem like this).

 

[lippyka]

b.) I'm fairly sure the magnitude is 0, however again, I can't seem to put the numbers together to show this. Any help would be greatly appreciated!

 

[baba89]

I understand your frustration with trying to solve this problem. It can be challenging to put all the pieces together and come up with the correct solution. However, it seems like you are on the right track with your attempt at solving for the electric field at the center of the ring.

To address your concerns, let's first review the concept of a 2-dimensional line charge. A 2-dimensional line charge is a distribution of charge along a line, in this case, the circumference of the circle. This means that the charge is spread out along the entire length of the line, and we need to integrate over this length to find the total electric field at a specific point.

Now, let's look at your approach. You correctly identified that we need to break down the electric field into its x and y components. However, the way you have done it is not quite correct. Let's take a closer look at your equations:

d\vec{E_{x}}=k\int\frac{dQ}{r^{2}}=k\int\frac{\lambda \;dL}{r^{2}}=k\int\frac{\lambda_{0}sin(\theta)\;dL}{r^{2}}

d\vec{E_{y}}=k\int\frac{dQ}{r^{2}}=k\int\frac{\lambda \;dL}{r^{2}}=k\int\frac{\lambda_{0}sin(\theta)\;dL}{r^{2}}

The first equation is correct, as it takes into account the total charge along the circumference, \lambda_{0}, and the distance from the center, r. However, the second equation is not quite right. The y-component of the electric field will depend on the distance from the center, r, and the angle, \theta, which is the angle between the line of charge and the y-axis. Therefore, we need to take into account the angle in our integration:

d\vec{E_{y}}=k\int\frac{dQ}{r^{2}}=k\int\frac{\lambda \;dL}{r^{2}}=k\int\frac{\lambda_{0}sin(\theta)\;dL}{r^{2}}sin\theta

This will give us the correct expression for the y-component of the electric field.

Now, to address your concern about