Field due to ring charge at x-y axis.

In summary, the electric field at any point on the positive z axis for a uniformly distributed charge Q around a thin ring of radius b is given by E = Qr / (b^2 + r^2)^3/2, where r is the distance from the point to the center of the ring. The maximum value for r is b / √2.
  • #1
Buffu
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Homework Statement



A charge ##Q## is distributed unifromly around a thin ring of radius ##b## which lies in ##xy## plane with its centre at the origin. Locate the point on the positive ##z## axis where the electric field is strongest ?

Physics ring charge.png


Homework Equations


The Attempt at a Solution



##\displaystyle d\vec E = {dQ \over b^2 + r^2 }\cos \alpha##. ##r## is the distance between the point, ##\alpha## is the angle between ##z## axis and the displacement vector , and the ring along ##z## axis.

##\displaystyle d\vec E = {\lambda dl \over b^2 + r^2 } \cos \alpha = d\vec E = {\lambda bd\theta \over b^2 + r^2 } \cos \alpha##.

Using ##\displaystyle \cos \alpha = {r \over \sqrt{b^2 + r^2}}##.

##\displaystyle E = {br\lambda \over (b^2 + r^2)^{3/2} } \int^{2\pi}_{0} d\theta = {2\pi br\lambda \over (b^2 + r^2)^{3/2} }##

Since ##\displaystyle \lambda = {Q\over 2\pi b}##,

I get,

##\displaystyle E = {Qr \over (b^2 + r^2)^{3/2}}##

Which has a maximum ##\displaystyle r = {2\over 3\sqrt{3}b^2}##.

Am I correct ?
 
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  • #2
The derivation of the electric field is correct. Your final answer is not. If b is the radius, your value for r has the wrong dimensions. It must proportional to b not to b-2.
 
  • #3
kuruman said:
The derivation of the electric field is correct. Your final answer is not. If b is the radius, your value for r has the wrong dimensions. It must proportional to b not to b-2.
Oh that was WA's mistake. I got lazy and let it calculate maxima. Is ## r = \pm b/ \sqrt{2}## correct ?
 
  • #4
Where is b on the right side of the equation?
 
  • #5
kuruman said:
Where is b on the right side of the equation?
sorry.

I edited my post. Is the answer ##\pm b/\sqrt{2}## ?
 
  • #6
I would pick the positive value because the problem specifies "on the positive z axis". Otherwise it's fine.
 
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Likes Buffu

1. What is the formula for calculating the electric field due to a ring charge at a point on the x-y axis?

The formula for calculating the electric field due to a ring charge at a point on the x-y axis is E = kqz / (z^2 + R^2)^(3/2), where k is the Coulomb's constant, q is the charge on the ring, z is the distance from the center of the ring to the point on the x-y axis, and R is the radius of the ring.

2. How does the distance from the center of the ring affect the electric field on the x-y axis?

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

3. Is the electric field on the x-y axis affected by the charge on the ring?

Yes, the electric field on the x-y axis is directly proportional to the charge on the ring. This means that as the charge increases, the electric field also increases.

4. What is the direction of the electric field on the x-y axis due to a ring charge?

The direction of the electric field on the x-y axis due to a ring charge depends on the location of the point on the axis. If the point is above the center of the ring, the field points downwards towards the negative y-axis. If the point is below the center of the ring, the field points upwards towards the positive y-axis.

5. Can the electric field on the x-y axis be negative?

Yes, the electric field on the x-y axis can be negative. This happens when the point on the axis is located inside the ring, as the electric field is directed towards the center of the ring, which is a negative charge. However, if the point is located outside the ring, the electric field will always be positive.

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