Electric field of a ring

In summary, the conversation discusses the electric field of a ring with radius r and distance z from the center, carrying a uniform line charge λ. It is mentioned that the radial component of the field cancels out due to symmetry and the question asks for a mathematical proof using cylindrical coordinate system. The answer involves converting the directional vector from the current element to the point of observation in cylindrical coordinates.
  • #1
rick_2009
1
0
This is about the electric field of a ring with radius r, at a distance z from center, along the axis of the ring. The ring carries a uniform line charge [tex]\lambda[/tex]. We always say that the radial component of the field cancels out due to symmetry. Can somebody tell how to prove it mathematically (using cylindrical coordinate system only)?

[tex]dE_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda d\theta}{(r^2+z^2)}\hat{\textbf{r}}[/tex]
[tex]E_{rad}=-\frac{1}{4\pi\epsilon_0}\frac{r\lambda}{(r^2+z^2)}\int_0^{2\pi} \hat{\textbf{r}} d\theta[/tex]

?
 
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  • #2
First thing you need to do is convert the directional vector that is local to the current element to a directional vector at the point of observation.

So the electric field element points in the r-hat direction, convert that to cylindrical coordinates and then also provide the appropriate translation since you will be observing it at a point offset from the origin of the coordinate system of the electric field element.
 
  • #3
I don't get your answer
 

What is an electric field?

An electric field is a region in space where an electrically charged particle experiences a force. This force is exerted by other charged particles in the field.

How is the electric field of a ring calculated?

The electric field of a ring can be calculated using the equation E = kq/r^2, where k is the Coulomb constant, q is the charge of the ring, and r is the distance from the center of the ring to the point where the electric field is being calculated.

What is the direction of the electric field around a ring?

The direction of the electric field around a ring is perpendicular to the plane of the ring, and it points away from the ring on one side and towards the ring on the other side.

How does the electric field of a ring change as the distance from the ring increases?

As the distance from the ring increases, the electric field decreases. This is because the electric field is inversely proportional to the square of the distance from the source of the field.

What is the relationship between the electric field and the charge of the ring?

The electric field is directly proportional to the charge of the ring. This means that as the charge of the ring increases, the electric field also increases.

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