Calculating the Electric field for a ring

In summary, the conversation discusses using the superposition principle to solve a problem involving two oppositely charged disks. The resulting charge distribution is an annulus and the total charge distribution should be normalized. There is a debate about whether using a double integral or a single integral is simpler, with the latter being the preferred approach due to symmetry arguments.
  • #1
link223
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Homework Statement
(II) A flat ring (inner radius R outer radius 4R ) is
uniformly charged. In terms of the total charge Q, determine the electric field on the axis at points (a) 0.25R and (b) 0.75R from the center of the ring. [Hint: The ring can be
replaced with two oppositely charged superposed disks.]
Relevant Equations
Gauss' Law
What i don't understand is why we are able to replace the ring with 'two oppositely charged superposed disks'?

Just trying to understand..
So we have a uniform charge which means that this'll just be a simplification of the problem than, correct?

Thanks in advance.
 
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  • #2
By the superposition principle, the solution to the sum of two charge distributions is the sum of the solutions to the separate charge distributions. If you take a small disk and a large disk with the opposite surface charge density (not charge!) then the sum of the charge distributions is an annulus as described in the problem as the resulting charge distribution within R is zero and between R and 4R there will only be the larger disc. Make sure to normalise your charge distribution such that the total charge distribution is of total charge Q.
 
  • #3
To me, the hint makes sense if the on-axis electric field due to a uniformly disk is assumed as given and there is no requirement to derive it. If there is such a requirement, I think doing the double integral would be simpler.
 
  • #4
kuruman said:
I think doing the double integral would be simpler.
Well, the double integral is also relatively easily written as a single integral just by symmetry arguments, making that approach even simpler. However, the difference in complexity is inserting zero twice in the primitive function (which evaluates to zero if you use the most convenient choice of integration constant).
 

1. How do you calculate the electric field for a ring?

To calculate the electric field for a ring, you can use the formula E = kqz/(z^2 + R^2)^(3/2), where k is the Coulomb's constant, q is the charge of the ring, z is the distance from the center of the ring, and R is the radius of the ring.

2. What is the direction of the electric field for a ring?

The electric field for a ring is always perpendicular to the plane of the ring and points away from the center of the ring for positive charges and towards the center for negative charges.

3. Can the electric field for a ring be negative?

Yes, the electric field for a ring can be negative if the charge of the ring is negative. This means that the electric field points towards the center of the ring instead of away from it.

4. How does the electric field for a ring change as you move away from the center?

The electric field for a ring decreases as you move away from the center. This is because the distance from the center (z) in the formula is in the denominator, meaning that as z increases, the electric field decreases.

5. Can the electric field for a ring be affected by other nearby charges?

Yes, the electric field for a ring can be affected by other nearby charges. This is because the electric field is a vector quantity and can be affected by the presence of other electric fields. The resulting electric field will be the vector sum of all the individual electric fields.

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