E field from half ring of charge.

In summary, the question is about finding the electric field at the center of a semi-ring of charge with total charge Q and radius R. The suggested method is to use Gauss's Theorem, but there is not enough symmetry for it to work. Instead, the direction of the E field can be determined by using cos\theta in the numerator of the integral. The charge density is represented by lambda = Q/2pi R.
  • #1
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Homework Statement


How do I find the electric field from a semi-ring of charge at the center of the would-be circle? Does that make sense? Or another way: If the semi-circle of charge is centered at the origin, what is the electric field at the origin? I hope you know what I mean now.

the total charge of the ring is Q, the radius is R.

Oh yeah, the semi circle is in the left hand plane

Homework Equations


[tex]
E = k\int_{}^{l}\frac{(r-r')\rho _l(r')}{|r-r'|^3} dl'

[/tex]

The Attempt at a Solution



[tex]
E = k\int_{\pi/2}^{-\pi/2}\frac{(-r')\rho _l(r')}{|-r'|^3} d\theta
[/tex]

Is converting to cylindrical coordinates the right move? Is this even the equation I should be using? What should I use as the charge density?
 
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  • #2
Try applying Gauss's Theorem. Assume a gaussian surface enclosing the ring and [tex]\int E.ds=\frac{q}{\epsilon}[/tex]
 
  • #3
Gauss won't work here because there is not full circular symmetry
There is enough symmetry to see the direction of the E field.
Just put cos\theta into the numerator of your integral.
rho is really a linear density lambda=Q/2pi R.
 

What is the formula for calculating the electric field from a half ring of charge?

The formula for the electric field from a half ring of charge is: E = (kQx)/(2pi((x^2 + R^2)^(3/2))), where k is the Coulomb's constant, Q is the charge of the ring, x is the distance from the center of the ring, and R is the radius of the ring.

How does the distance from the center of the ring affect the electric field?

The electric field is inversely proportional to the distance from the center of the ring. This means that as the distance increases, the electric field decreases.

Is the electric field from a half ring of charge uniform?

No, the electric field from a half ring of charge is not uniform. The electric field is stronger at points closer to the ring and weaker at points further away. This is due to the inverse square law.

How does the charge of the ring affect the electric field?

The electric field is directly proportional to the charge of the ring. This means that as the charge increases, the electric field also increases. However, the distance from the ring also affects the strength of the electric field.

Can the electric field from a half ring of charge be negative?

Yes, the electric field from a half ring of charge can be negative. This happens when the point of interest is on the opposite side of the ring compared to the direction of the electric field. In other words, the electric field will be negative when x is greater than R in the formula.

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