# Generating Expressions For Electric Flux

1. Feb 9, 2013

### Bashyboy

1. The problem statement, all variables and given/known data
A particle with charge Q is located immediately above the center of the flat face of a hemisphere of radius R as shown in the figure below.

(a) What is the electric flux through the curved surface? (Use any variable or symbol stated above along with the following as necessary: ε0.)

(b) What is the electric flux through the flat face? (Use any variable or symbol stated above along with the following as necessary: ε0.)

2. Relevant equations

3. The attempt at a solution
Taking the surface area equation for a sphere, and dividing by two: $2 \pi r^2$

Area equation for a circle: $\pi r^2$

The expression for (a): $\Phi = E2 \pi r^2$

The expression for (b): $\Phi = E \pi r^2$

These are incorrect. What did I do wrong?

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Last edited: Feb 9, 2013
2. Feb 9, 2013

### TSny

You didn't express the answers in terms of the symbols required. Part (b) doesn't look correct.

3. Feb 9, 2013

### Bashyboy

Well, what other relationships should I use, in order to put it in the form that question asks for?

4. Feb 9, 2013

### TSny

What does E stand for in your answers? What produces E? Can you express E in terms of the symbols Q, R, and εo?

5. Feb 9, 2013

### Bashyboy

Oh, so for (a): $\Phi = \large \frac{Q}{2 \epsilon_0}$ Which would make sense, seeing as if the object were a sphere, and it enclosed the charge at its center, the whole sphere would experience an electric flux of $\Phi = \large \frac{Q}{\epsilon_0}$; but since it is half a sphere, half of the of electric field is emanating away from the bottom half of this sphere.

6. Feb 9, 2013

### TSny

Yes, that looks correct.

7. Feb 9, 2013

### Bashyboy

You said that my answer for part (b) didn't look correct, care to elaborate? I'd appreciate it.

8. Feb 9, 2013

### TSny

In general the flux through an area is not simply the product of the magnitude of E and the area. That only works if the field is uniform over the surface and perpendicular to the surface. Neither of these conditions is met for the flat face.

9. Feb 9, 2013

### Bashyboy

The field isn't parallel to the area vector at any point on the circular surface? How am I to solve this part, then?

10. Feb 9, 2013

### TSny

Think Gauss' law which applies to any closed surface.

11. Feb 9, 2013

### Bashyboy

This isn't a closed surface, though.

12. Feb 9, 2013

### TSny

Right. The flat surface by itself is not a closed surface. But is it part of a closed surface?

13. Feb 9, 2013

### Bashyboy

Yes. What does that imply?

14. Feb 9, 2013

### TSny

What does Gauss' law tell you about the total flux through this particular closed surface?

15. Feb 9, 2013

### Bashyboy

Does it have to do with the amount of electric fields going into the surface equals the amount exiting? Meaning the flux is the same for both surfaces? Why is this true?

16. Feb 9, 2013

### TSny

Can you state Gauss' law?

17. Feb 9, 2013

### Bashyboy

I looked at Wikipedia's definition, but I am unable to state it in my own words.

18. Feb 9, 2013

### TSny

Wkipedia says "The electric flux through any closed surface is proportional to the enclosed electric charge". How much charge is enclosed in your closed surface?

19. Feb 9, 2013

### Bashyboy

There isn't any charge within the closed surface; only electric fields pass through the surface.

Last edited: Feb 9, 2013
20. Feb 9, 2013

### TSny

That's right. Since Gauss' law states that the total flux through the closed surface is proportional to the charge within, what can you deduce about the value of the total flux?