# Question 1

1. Feb 4, 2005

### kyang002

What is the electric flux through the surface shown in the figure (attachment)?

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2. Feb 4, 2005

### arildno

What ideas do you have?

3. Feb 4, 2005

### kyang002

I'm unsure of how to do this problem. No ideas so far.

4. Feb 4, 2005

### arildno

What is the definition of electric flux?
(Look up in your book)

And, do NOT double post (posting the same question twice or more), and show your own work or at least your ideas/what you are uncertain of in the future.

5. Feb 4, 2005

### Mk

Well, of course as he/she said: he/she has no ideas so far

6. Feb 4, 2005

### ahrkron

Staff Emeritus
Just saying "I have no idea" is like asking to NOT be helped. At least, for sure, kyang002 can find the definition of electric flux, and then try at least to correlate it with the problem.

Actually, just typing the definition here will provide a starting point for others to help.

7. Feb 4, 2005

### K.J.Healey

I think writing down the equation for electric flux would just about answer this question completely....(after plugging in 2 numbers)

8. Feb 4, 2005

### kyang002

Electric Flux would be E x A or EA cos theta.

9. Feb 4, 2005

### kyang002

The answer is 1. But I have no idea of doing it. Again no help is given.

10. Feb 4, 2005

### dextercioby

I simply hope u didn't mean
$$|\vec{E}\times\vec{A}|=EA\cos\theta$$

That "x" would invite to an erroneous interpretation.

To the OP:It's only about applying the (simple) definition and using a bit of trigonometry...

Daniel.

11. Feb 4, 2005

### K.J.Healey

E*A*Cos(theta)

Ok, they give you E,A,and theta. PLUG IT IN A CALCULATOR

You see, flux is the measurement of the field passing through the surface, times the area of the surface (simplest, non integral way).

So if your field is at a 30 degree angle to the horizontal surface, you need whats passing perpindicularly(right-angle) to the surface.

So its a right triangle,
/|
/ |
----

And if you want the Y edge, Cos(left angle) = Height/Hypot
Hypot = E field so
Efield*(cos(theta)) = Height

So E*Cos(theta) * Area = fllux

12. Feb 4, 2005

### arildno

kyang:
1. Could you identify the quantities you have been given?
2. Where should those quantities be plugged into the formula you have presented?
(That is, the flux is given by $$EA\cos\theta$$)

Note: Be careful with the angle!

13. Feb 4, 2005

### kyang002

Now I understand. Thanks guys.