Gauss' Law: Sphere with an Opening

[Const@ntine]

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

Φ_Ε = ∫E^→⋅dA^→ (for a surface)

Φ_Ε = q_internal/ε₀ (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 Φ_Ε = 28.2 Nm²/C

 

[Orodruin]

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.

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.

 

[Const@ntine]

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.

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 Φ_Ε = ∫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.

 

[Orodruin]

E*A = E*4*π*r2

What is the area of a circle?

 

[Const@ntine]

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!