Electric Field of a Half Charged Circle

In summary: The final expression is correct. In summary, the electric field at a distance away (a) that's in the middle of the circle is equal to 2 times the integral of (1/4πεo) * (r/√(a^2+r^2)) * (δπr/(a^2+r^2)), integrated from r = 0 to r = R.
  • #1
glueball8
346
1

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


[tex] E=\frac{1}{4 \pi \epsilon_{o}} \times \frac{Q}{R^2} [/tex]

The Attempt at a Solution



So I got til here.

[tex] 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 [/tex]

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

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

Thanks.
 
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  • #2
In your expression of dE where is the charge?
 
  • #3
rl.bhat said:
In your expression of dE where is the charge?

ops :blushing:

[tex] 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
[/tex]
 
  • #4
Bright Wang said:
ops :blushing:

[tex] 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
[/tex]
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.
 
  • #5
rl.bhat said:
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
 
  • #6
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.
 
  • #7
rl.bhat said:
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?

[tex] 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} [/tex]

that's the solution?
 
  • #8
Yes.
 

1. What is the definition of electric field?

The electric field is a physical quantity that describes the strength and direction of the force experienced by a charged particle placed in the field.

2. How is the electric field of a half charged circle calculated?

The electric field of a half charged circle can be calculated using the formula E = (k*q)/r^2, where k is the Coulomb's constant, q is the charge on the circle, and r is the distance from the center of the circle to the point where the electric field is being measured.

3. What is the direction of the electric field of a half charged circle?

The direction of the electric field of a half charged circle is perpendicular to the surface of the circle at every point. This means that the electric field lines will point away from the positively charged half and towards the negatively charged half of the circle.

4. How does the electric field of a half charged circle vary with distance?

The electric field of a half charged circle follows an inverse-square law, meaning that as the distance from the circle increases, the electric field strength decreases. This is because the electric field lines spread out as they move away from the circle.

5. Can the electric field of a half charged circle be zero?

Yes, the electric field of a half charged circle can be zero at certain points. This occurs when the distance from the center of the circle is equal to the radius of the circle. At this point, the electric field contributions from the positive and negative halves of the circle cancel each other out.

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