Calculating Flux Through a Circular Ring

[Tanya Sharma]

Homework Statement

A particle having charge q = 8.85 μC is placed on the axis of a circular ring of radius R = 30 cm. Distance of the particle from centre of the ring is a = 40 cm. Calculate electrical flux passing through the ring.

Homework Equations

Flux through a surface = ∫E.ds

The Attempt at a Solution

I have tried to sketch a figure .

Since the circle is not a closed surface .I tried to make the spherical cap whose base is the given circle . Then I completed that sphere .

Now I think flux through the circle is equal to the flux through the spherical cap .But then what is the flux through the spherical cap ?

Is the ratio of the surface area of the cap to that of the sphere same as the ratio of the flux across these two surfaces ?

I would appreciate if somebody could help me with the problem .

 

[ehild]

Homework Statement

A particle having charge q = 8.85 μC is placed on the axis of a circular ring of radius R = 30 cm. Distance of the particle from centre of the ring is a = 40 cm. Calculate electrical flux passing through the ring.

Homework Equations

Flux through a closed surface = q/ε

The Attempt at a Solution

I have tried to sketch a figure .

Since the circle is not a closed surface .I tried to make the spherical cap whose base is the given circle . Then I completed that sphere .

Now I think flux through the circle is equal to the flux through the spherical cap .But then what is the flux through the spherical cap ?

Is the ratio of the surface area of the cap to that of the sphere same as the ratio of the flux across these two surfaces ?

I would appreciate if somebody could help me with the problem .

It is possible to solve the problem in your way, using the area of the spherical cap.
For definition of the electric flux, read http://en.wikipedia.org/wiki/Electric_flux

ehild

 

[Tanya Sharma]

Sorry..i do not understand what you are trying to convey .

Is there some other way to solve the problem ? All I could think of is what I have written in post#1 .

My understanding of electric flux is simply ∫E.ds for a surface .

Could you reflect more on how to approach this problem ?

 

[ehild]

You wanted to find the flux across the circle enclosed by the ring by the flux through the cap.
You can determine the flux from its definition, ∫E.ds directly. You know the electric field on the plane of the ring, both magnitude and the angle it encloses with the normal of the plane.

ehild

 

[Tanya Sharma]

But the direction as well as magnitude of Electric field is different for different points on the circular region .It will be same for points equidistant from the center of the circle.

 

[Saitama]

Now I think flux through the circle is equal to the flux through the spherical cap .But then what is the flux through the spherical cap ?

Hi Tanya!

You can find flux through a sphere, right? If you select a cap of area A, can you determine the flux through A (in terms of A)? Can you then express A in the given quantities?

 

[ehild]

But the direction as well as magnitude of Electric field is different for different points on the circular region .It will be same for points equidistant from the center of the circle.

Both methods are correct. If you know the area of the cap, that method is easier. And yes, the ratio of the fluxes is equal to the ratio of the areas, that is, the surface area of the cap to the surface of the sphere.
If I were you, I solved the problem in both ways, just for fun.

ehild

 

[voko]

But the direction as well as magnitude of Electric field is different for different points on the circular region .It will be same for points equidistant from the center of the circle.

You need to understand that when you have a closed contour, and two surfaces that are "stretched" onto that contour, the electric flux through the two surfaces is identical if there is no charge between the surfaces. This follows directly from Gauss's law.

That is why the problem talks about flux through the contour without specifying any particular surface. Choose the surface that makes it easiest to compute the flux.

 

[Tanya Sharma]

Hi Tanya!

You can find flux through a sphere, right? If you select a cap of area A, can you determine the flux through A (in terms of A)? Can you then express A in the given quantities?

Hello Pranav...

Thanks for the input.

I haven't dealt with spherical caps before .

Is the surface area $A = 2\pi(a^2+d^2)(1-\frac{d}{\sqrt{a^2+d^2}})$ where 'a is the radius of the circle and 'd' is the distance between the charge 'q' and the center of the circle.

$Flux = \int \vec{E} \cdot \vec{ds} = EA = \frac{q}{2ε_0}(1-\frac{d}{\sqrt{a^2+d^2}})$

Does this makes sense ?

 

[Saitama]

Hello Pranav...

Thanks for the input.

I haven't dealt with spherical caps before .

Is the surface area $A = 2\pi(a^2+d^2)(1-\frac{d}{\sqrt{a^2+d^2}})$ where 'a is the radius of the circle and 'd' is the distance between the charge 'q' and the center of the circle.

$Flux = \int \vec{E} \cdot \vec{ds} = EA = \frac{q}{2ε_0}(1-\frac{d}{\sqrt{a^2+d^2}})$

Does this makes sense ?

Yes, it is correct. :)