Electric flux through a 1.00 mm hole drilled in a 10.0 cm sphere

[Pratik007789]

Homework Statement

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An uncharged nonconductive hollow sphere of radius 10.0 cm surrounds a 20.0 µC charge located at the origin of a cartesian coordinate system. A drill with a radius of 1.00 mm is aligned along the z axis, and a hole is drilled in the sphere. Calculate the electric flux through the hole.

Homework Equations

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Volume charge distribution = Q/V
Flux = q/ε0 = EA

The Attempt at a Solution

In the picture
But the answer is coming out to be wrong

 

[berkeman]

Could you please try to Upload your work again so that it is not sideways? Thanks.

Also, I didn't look at your work yet (since it was sideways), but are the dielectric constant and the thickness of the non-conducting sphere given?

 

[kuruman]

I looked at your work and I was able to read it because I have a laptop that I can rotate 90^o. Your method is correct, you multiply the total flux Q/ε₀ by the ratio of the areas of hole to sphere. The only problem is that the area of the sphere is not πr². If you use the correct expression for the area, you will be OK.

 

[Pratik007789]

I looked at your work and I was able to read it because I have a laptop that I can rotate 90^o. Your method is correct, you multiply the total flux Q/ε₀ by the ratio of the areas of hole to sphere. The only problem is that the area of the sphere is not πr². If you use the correct expression for the area, you will be OK.

But If I we have to ratio at the end then the other terms like 2, π won't cancel out? Any support would be appreciated

 

[haruspex]

But If I we have to ratio at the end then the other terms like 2, π won't cancel out? Any support would be appreciated

Not sure what you mean, but I feel you have not understood kuruman's response. What is the surface area of a sphere radius r? Use that instead of the πr² that you used.

 

[kuruman]

What is the surface area of a sphere radius r? Use that instead of the πr2 that you used.

Yes, do that and if something cancels out, so be it. You will have a simpler expression to deal with.

 

[Pratik007789]

Thanks guys. Really appreciate your help and support.