Electric Field of a Half Charged Circle

[glueball8]

Homework Statement

I'll try to describe. There's a circle with radius R and a charge density of positive on the top half of the circle and a charge density of negative on the bottom. What is the electric field at a distance away (a) that's in the middle of the circle?

Homework Equations

$$E=\frac{1}{4 \pi \epsilon_{o}} \times \frac{Q}{R^2}$$

The Attempt at a Solution

So I got til here.

$$dE=\frac{1}{4 \pi \epsilon_{o}} \times \frac{r}{\sqrt{a^2+r^2}} \times \frac{\pi*r*dr}{a^2+r^2} * X$$

I don't know how to account for the y component of r. It have to times a multiple, X. Is X $$\frac{1}{\sqrt{2}}$$? Just a guess.

Then I can take the integral after. Any tip on how to do X?

Thanks.

 

[rl.bhat]

In your expression of dE where is the charge?

 

[glueball8]

In your expression of dE where is the charge?

ops

$$dE=\frac{1}{4 \pi \epsilon_{o}} \times \frac{r}{\sqrt{a^2+r^2}} \times \frac{\delta*\pi*r*dr}{a^2+r^2} * X$$

 

[rl.bhat]

ops

$$dE=\frac{1}{4 \pi \epsilon_{o}} \times \frac{r}{\sqrt{a^2+r^2}} \times \frac{\delta*\pi*r*dr}{a^2+r^2} * X$$

x-components of two semicircular charges add up and y-components cancel. When you take integration the total charge will be δπr. Due to two semicircular charges X should be 2.

 

[glueball8]

x-components of two semicircular charges add up and y-components cancel. When you take integration the total charge will be δπr. Due to two semicircular charges X should be 2.

I don`t understand. I will do the integral over the whole area. The net force will be downwards because of symmetry. But for most points the x component is cancel by the other side. so I need to do something to take that into account.

Thanks

 

[rl.bhat]

the electric field at a distance away (a) that's in the middle of the circle?
Do you mean along the axis of the circle?
Consider a thin ring of the charged circular plate.. Two small segments of the ring situated diametrically opposite each other at a point P at a distance a from the center, will have field dE, one moving away from the positive portion of the ring and another towards the ring due to negative side of the ring. If you resolve these two into X and Y components, Y components get canceled out and X-components add up.
Your expression for dE in post #3 is correct except the term X. Integrate it between the limits r = 0 to r = R. And double it to take in acount the negative half circle.

 

[glueball8]

the electric field at a distance away (a) that's in the middle of the circle?
Do you mean along the axis of the circle?
Consider a thin ring of the charged circular plate.. Two small segments of the ring situated diametrically opposite each other at a point P at a distance a from the center, will have field dE, one moving away from the positive portion of the ring and another towards the ring due to negative side of the ring. If you resolve these two into X and Y components, Y components get canceled out and X-components add up.
Your expression for dE in post #3 is correct except the term X. Integrate it between the limits r = 0 to r = R. And double it to take in acount the negative half circle.

Ok, so X is just 1?

$$E=2 \int \frac{1}{4 \pi \epsilon_{o}} \times \frac{r}{\sqrt{a^2+r^2}} \times \frac{\delta*\pi*r*dr}{a^2+r^2}$$

that's the solution?

 

[rl.bhat]

Yes.