Electric Flux Through the Surface of a Circle

In summary: So you are left with ##\Omega## which you can calculate using the angle subtended by the cap at the center of the sphere.If you are not familiar with solid angles, you can look up the formula for the solid angle of a cone in terms of its apex angle.When using this method, you don't have to do any integration. Just find the total charge using the charge density and then use the solid angle to get the flux through the disk.In summary, the problem is to find the flux of an electric field through a circular surface tangent to a charged sphere with a given charge density. Using Gauss' law and the concept of total charge enclosed, the flux is calculated by integrating the magnitude of the electric field
  • #1
TerraForce469
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0

Homework Statement



A sphere of radius ## R ## carries charge density ## \rho = ar^5 ## where ## a ## is a constant. Find the flux ## \Phi ## of its electric field through a surface of a circle with radius ## R ## if the circle lies in a plane tangent to the sphere and its center touches the sphere.

Homework Equations



Gauss' law to calculate ## \vec{E} ## of the charged sphere: $$\oint \vec{E}\cdot d\vec{A} = \frac {Q_{enc}}{ε_0}$$

The total charge of the sphere (assuming volumetric charge density, wasn't specified which type in the original prob.): $$ Q_{enc} = \int_V \rho(r)\,dV $$

And of course, the electric flux: $$ \Phi_E = \int \vec{E}\cdot d\vec{A} $$

The Attempt at a Solution



I found total charge enclosed by integrating all over the volume of the sphere and obtained: $$ Q_{enc} = \frac{1}{2} \pi a R^8 $$

Then I used Gauss' law to find ## \vec{E} ## and obtained: $$ \vec{E} = \frac{aR^8}{8ε_0r^2} \hat{r} $$

The surface over which I wanted to find the flux, take it to be at the right of the sphere and touching it at its center. I will call its differential area ## dA' = r' d r' d \theta ##.

The only contributions of ##\vec{E}## come from its component normal to ##\vec{dA'}##. Thus we need to find the magnitude corresponding to that component using: $$ cos(α) = \frac{R}{r} $$ where ## r= \sqrt{r'^2 + R^2} ##. I apologize if you cannot visualize it or understand how I am approaching this, but bear with me. Basically α is the angle between the radius ## R ## of the sphere and ## r ##, the distance of the ## \vec{E} ## component.

So now I take the flux:

$$ \Phi_E = \int_S \vec{E}\cdot d\vec{A'} = \int_S E cos(α) dA' $$

And then substitute in my previous expressions for ## E ## and ## cos(α) ##, and integrating all over the flux surface:

$$ \Phi_E = \int_0^{2\pi} \int_0^R (\frac{aR^8}{8ε_0r^2})(\frac{R}{r}) r' dr' d \theta' = \int_0^{2\pi} \int_0^R (\frac{aR^8}{8ε_0(r'^2 + R^2)})(\frac{R}{\sqrt{r'^2 + R^2}}) r' dr' d \theta' $$

(I don't know how to write the equation in steps using Latex, if someone could please inform me how to on the forums that would be appreciated). My final solution is:

$$ \Phi_E = \frac{\pi a R^8}{4ε_0} (1 - \frac{1}{\sqrt{2}}) $$

I was hoping to check if my methodology was correct and if I am doing something wrong. This problem took me a while to do and I want to at least check that it seems proper. Feel free to ask and I will clarify. Thank you!
 
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  • #2
Hi TerraForce469.

Your work looks correct to me. Nicely done!
 
  • #3
TSny said:
Hi TerraForce469.

Your work looks correct to me. Nicely done!

Hi TSny! :smile:

I was wondering if there is any elegant solution to this problem. I am not sure but can we use the spherical cap method you taught me recently?

EDIT: That seems to work, I should have tried it before posting the reply. :redface:
 
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  • #4
Hi Pranav-Arora.

Yes! TerraForce469's write-up was so nicely done that I didn't even think about another way.

You can easily check that it will yield the correct answer without doing any flux integration (if you know or look up the formula for the solid angle of a cone in terms of its apex angle.) Of course you still need to do the integral to get the total charge.

Thanks for reminding me.
 
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  • #5
Terraforce469,

If you want to try Pranav-Arora's suggestion, then extend the radial lines from the center of the sphere through the edge of the disk to produce a cone. Then cap the cone with a spherical cap that is part of a sphere centered on the center of the charged sphere. Compare the flux through the disk with the flux through the spherical cap.

[You don't need to extend the radial lines, you could just cap the disk with a spherical cap.]
 

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  • #6
Pranav-Arora and TSny,

Sorry I couldn't respond any sooner; quite a busy day I had today.

Thank you for your responses! Although we are supposedly still at rudimentary work, it's reassuring to know that the solution to a difficult problem is correct when I good critical reasoning of fundamental relevant concepts are applied.

At a glance, the spherical cap method does seems more intuitive. I can see how the flux through the spherical cap is equivalent to that of the disk. However, what changes from my original approach? Is that the solid angle of the spherical cap is taken into account?
 
  • #7
TerraForce469 said:
At a glance, the spherical cap method does seems more intuitive. I can see how the flux through the spherical cap is equivalent to that of the disk. However, what changes from my original approach? Is that the solid angle of the spherical cap is taken into account?

Yes. The field is radial and has uniform magnitude over the surface of a sphere centered on the charge distribution. So, the flux through a spherical cap is just the magnitude of the field times the area of the cap.

The area of the cap is ##r^2\Omega## where ##\Omega## is the solid angle subtended by the cap.

##r^2## cancels out when multiplying the area by the field since E varies as ##1/r^2## outside the charge distribution.
 

1. What is electric flux through the surface of a circle?

Electric flux through the surface of a circle is the measure of the total amount of electric field passing through a closed surface in the shape of a circle. It is a scalar quantity and is measured in units of volts.

2. How is electric flux through the surface of a circle calculated?

The electric flux through the surface of a circle can be calculated by multiplying the strength of the electric field at each point on the surface by the area of the surface and then taking the dot product of these values. This value represents the total electric flux passing through the surface.

3. What factors affect the electric flux through the surface of a circle?

The electric flux through the surface of a circle is affected by the strength of the electric field, the size of the circle, and the angle between the electric field and the surface. A larger surface area or stronger electric field will result in a higher electric flux, while a smaller angle between the electric field and surface will result in a lower electric flux.

4. Why is electric flux through the surface of a circle important?

Electric flux through the surface of a circle is an important concept in electromagnetism as it helps us understand the behavior of electric fields. It is used to calculate the amount of electric field passing through a given surface and is crucial in many applications such as designing electrical circuits and understanding the behavior of charged particles.

5. How does the direction of the electric field affect the electric flux through the surface of a circle?

The direction of the electric field is an important factor in calculating electric flux through the surface of a circle. If the electric field is perpendicular to the surface, the electric flux will be at its maximum value. However, if the electric field is parallel to the surface, the electric flux will be zero as no electric field lines are passing through the surface.

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