# Need Help in Finding Electric Flux through a Hole in Sphere

1. Oct 2, 2011

### twisted079

1. The problem statement, all variables and given/known data

An uncharged nonconductive hollow sphere of radius 19.0 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.

2. Relevant equations

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

3. The attempt at a solution

So I started off by finding the Volume which was (4/3)pi(0.19^3) = 0.0287
I then divided the charge by this to get the Volume Charge Density = 6.9686^-4

Here is where I get stuck. I have no idea how to relate the Volume Charge Density to the flux through the small hole. I thought perhaps I should find the Area Charge Density of the small hole but I am really unsure. Im not asking for the answer, I just need a brief description of where to go from here/ what formulas to use. Any help would be greatly appreciated!

2. Oct 2, 2011

### Staff: Mentor

Rather than looking at charge density for the volume, start with the flux coming through the entire surface of the sphere. What portion of it comes through the small hole?

3. Oct 2, 2011

### twisted079

19.0 cm = 0.19 m
1.0 mm = .001 m

I suppose the ratio in terms of radius would be 0.001 : 0.19.
0.001 / 0.19 = .0052

Would this be the percent of the entire flux which passes though the hole?

4. Oct 2, 2011

### Staff: Mentor

You'll want to look at the ratio of the areas.

5. Oct 2, 2011

### twisted079

So I would calculate the surface areas of both the hole and the sphere, and use the ratio to calculate the flux?

6. Oct 2, 2011

### Staff: Mentor

The flux leaves the sphere isotropically -- it's uniform over all the surface of the sphere. So the bit leaving the area of the hole is proportional to the area of the hole. The ratio of the area of the hole to the area of the sphere is the proportion of the total flux that leaves by the hole. So yes... multiply the ratio by the total flux and you're done.

7. Oct 2, 2011

### twisted079

Awesome! I cant thank you enough for the help, especially since it is coming from a stranger who is truly kind enough to help someone he/she does not know.

8. Oct 2, 2011

### Staff: Mentor

You're quite welcome

9. Oct 2, 2011

### twisted079

Ah! Ive been trying this for quite some time and I still cannot figure it out.

I use 4piR2(Q/4piε0R2) to find the flux on the surface. I then multiply it by the ratio like you said but get the wrong answer :(

What am I doing wrong?

10. Oct 2, 2011

### SammyS

Staff Emeritus
What do you get for the two areas that you're taking the ratio of?

11. Oct 2, 2011

### twisted079

The areas would have to be:
1) 4pi(.192) = .4536
2) 4pi(.0012) = 1.256e-5

12. Oct 2, 2011

### SammyS

Staff Emeritus
The area of a circle (circular hole) is not A = 4πr2. That's 4 times the area !

13. Oct 2, 2011

### twisted079

*face palm*
Why no, no its not.
Its funny how such a simple error can cause so much frustration.
Now my calculations are coming out correct. Thank you!

14. Oct 2, 2011

### iRaid

Hated it when that happened lol.. At first your like, no way... Did I really do that :p