Electric field of a charged ring

In summary, the conversation discusses finding the magnitude of the electric field along the positive z axis using previous work on the problem. The solution involves integrating around the ring with respect to q, but since all of the ring charge is equally distant from the point on the z axis, calculus is not necessary and the final expression is E = kq/r^2 * (z/(sqrt(a^2+z^2)) where r^2 = (a^2+z^2).
  • #1
Seraph404
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0

Homework Statement



This is a http://session.masteringphysics.com/problemAsset/1003223/45/14677_a.jpg" of the problem

-I know that the direction of the electric field at any point on the z-axis is parallel to the z-axis
E = k*q/r^2

http://session.masteringphysics.com/problemAsset/1003223/45/1003223B2.jpg"

dE = k* dq/(a^2+z^2)

dEz = (z/(sqrt(a^2+z^2)) *dE

The Attempt at a Solution



So, I need to find the magnitude of the electric field along the positive z axis.

Basically, everything up there was work from previous parts of the problem. Now, I think I just need to put the last two expressions together in order to integrate around the ring (I think with respect to q), but I'm not sure. I need a little help on this last step.
 
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  • #2
I don't know if there's a better way to explain.
 
  • #3
dE = k* dq/(a^2+z^2)
dEz = (z/(sqrt(a^2+z^2)) *dE
Those look good! When you integrate around the ring, a,z and k are constant so you'll have something times Integral of dq from 0 to q, which is just the total charge on the ring, q. The integral just finds the total charge, which you already know anyway!

You don't need calculus for this one because all of the ring charge is equally distant from the point on the z axis where you are calculating and you want only the z component of E.
You just get E = kq/r^2 * (z/(sqrt(a^2+z^2)) where r^2 = (a^2+z^2).
 
Last edited:

Related to Electric field of a charged ring

1. What is an electric field?

An electric field is a physical quantity that describes the influence that a charged object has on other charged objects. It is a vector quantity, meaning it has both magnitude and direction.

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

The electric field of a charged ring can be calculated by dividing the charge on the ring by the distance from the ring to the point where the electric field is being measured. This calculation takes into account the size and shape of the ring, as well as the charge it carries.

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

The electric field of a charged ring points away from the ring if the charge is positive, and towards the ring if the charge is negative. The direction of the electric field can be determined by using the right-hand rule, where the thumb points in the direction of the charge and the fingers curl in the direction of the electric field.

4. How does the distance from the charged ring affect the electric field?

The electric field of a charged ring decreases as the distance from the ring increases. This is because the electric field follows an inverse square law, meaning it decreases with the square of the distance from the source.

5. How is the electric field affected by the size of the charged ring?

The electric field of a charged ring is directly proportional to the size of the ring. This means that as the size of the ring increases, the electric field also increases. However, the distance from the ring also plays a role, as the electric field decreases with distance, so a larger ring may have a stronger electric field closer to it, but a weaker electric field farther away.

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