Gauss' Law: Sphere with an Opening

In summary, using Gauss' Law, an electric flux of 28.2 Nm2/C is flowing through the opening in the sphere surrounding the electric charge.
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Const@ntine
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Homework Statement


An uncharged, unconductive, hollow sphere with a radius R of 10.0 cm, surrounds an electric charge of 10.0 μC, which is found at the beginning of the axises, in a standard cartesian system.

Parallel to the z axis, a small drill with a radius r = 1.00 mm opens a hole in the sphere.

What's the Electric Flux that goes through that opening?

Homework Equations


ΦΕ = ∫E⋅dA (for a surface)

ΦΕ = qinternal0 (Gauss' Law)

The Attempt at a Solution


I'm honestly pretty lost here. The book doesn't offer any examples, and only has a small paragraph on this part. From what I'm getting, I need to find the EF on that single part, so I cannot use the Law, since it refers to a whole area that surrounds the charge.

At first, I figured that since the charge is in the beginning of the axises, then its distance from the opening would be: d = R - 2r = 0.098 m

Problem is, I get stuck there because I'm not exactly sure how the whole thing works yet. The integration for example. The book says that E is always constant on the surface. Because its vector is parallel to the surfaces, the angle between them is 0. And so you have only dA to integrate, which results in just the A. I tried doing that computation (E*A) but I don't get the correct result.

Any help untangling this so that I can understand the basics would be appreciated!

PS: The book's answer is ΦΕ = 28.2 Nm2/C
 
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Techno_Knight said:
At first, I figured that since the charge is in the beginning of the axises, then its distance from the opening would be: d = R - 2r = 0.098 m
The hole is at the surface. The distance from the surface to the charge is always 10 cm.

Techno_Knight said:
And so you have only dA to integrate, which results in just the A. I tried doing that computation (E*A) but I don't get the correct result.
Please show us how you did this computation. Otherwise it is impossible for us to discover where you go wrong.

I get the same result as your book.
 
  • #3
Orodruin said:
The hole is at the surface. The distance from the surface to the charge is always 10 cm.
Ah, so it's just a circle, not a sphere. The opening, I mean. I just assumed that since drills always a spherical-esque end, the opening would be create in all 3 axises, xyz, not just xy. It'd have depth I mean.

Orodruin said:
Please show us how you did this computation. Otherwise it is impossible for us to discover where you go wrong.
I get the same result as your book.

Well, we know that E = kq/R2, which is the Electrical Field in every part of the surface of the sphere. With k = 8.99*109 Nm2/C2, q & R from the exercise's data, we get E = 8990000 N/C

Then, we have ΦΕ = ∫E⋅dA = ∫Ecos0dA = E*∫dA = E*A = E*4*π*r2 = 113 Nm2/C

Which obviously is wrong, but I don't know what I'm doing wrong.
 
  • #4
Techno_Knight said:
E*A = E*4*π*r2
What is the area of a circle?
 
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  • #5
Orodruin said:
What is the area of a circle?
...Yeah, I uh... This is embarassing... It's π*r2, not 4*π*r2, that's for a sphere. Using that I get the proper result.

Thanks for the help!
 

FAQ: Gauss' Law: Sphere with an Opening

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface.

2. How does Gauss' Law apply to a sphere with an opening?

Gauss' Law can be used to determine the electric field at any point outside or inside a sphere with an opening by considering the charge enclosed by the closed surface.

3. What is the equation for Gauss' Law?

The equation for Gauss' Law is ∮S E ∙ dA = Qenc / ε0, where ∮S E ∙ dA represents the electric flux through a closed surface, Qenc is the total charge enclosed by the surface, and ε0 is the permittivity of free space.

4. How do you apply Gauss' Law to solve problems involving a sphere with an opening?

To apply Gauss' Law to a sphere with an opening, you must first determine the charge enclosed by the surface, then use the equation ∮S E ∙ dA = Qenc / ε0 to solve for the electric field at the desired point.

5. What are some real-world applications of Gauss' Law?

Gauss' Law has many real-world applications, such as calculating the electric field around a charged spherical object, determining the electric potential inside a hollow conducting sphere, and analyzing the electric field between two oppositely charged parallel plates. It is also used in the design of electronic devices and power transmission systems.

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