Calculate the flux of the indicated electric field vector through the circle

In summary, the flux of the electric field vector through a surface is dependent on the amount of charge within the surface, and the surface's geometry.
  • #1
cwatki14
57
0
Calculate the flux of the indicated electric field vector through the surface. (E = 130, = 66.0°.)
p16-68alt.gif
I know the differential equation for flux d[tex]\phi[/tex]= A [tex]\bullet[/tex] d S
[/B]

My textbook told me that "if there is zero total charge within the closed surface S, there is no net flux of the electric field vector through S."
I guessed that the total flux was 0, but that was wrong...
Or maybe it's
[tex]\int[/tex] (from O to .05m) <130,0> [tex]\bullet[/tex] [tex]\pi[/tex] r ^2(<cos60,sin60>)
But I'm not really even sure how to do an integral with a dot product(would you dot them first?), and I'm fairly certain that's wrong...

Any ideas?
 
Last edited:
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  • #2
Welcome to PF!

Hi cwatki14! Welcome to PF! :smile:

(type \cdot, not \bullet :wink:)
cwatki14 said:
My textbook told me that "if there is zero total charge within the closed surface S, there is no net flux of the electric field vector through S."
I guessed that the total flux was 0, but that was wrong...

A "closed surface" mans a surface with no boundary (like a sphere), but S does have a boundary … it's the circle round S :wink:

(I suspect you're confusing this with a topologically closed (as opposed to open) surface … a surface which includes its boundary)
 
  • #3
So is the flux then
[tex]\int[/tex](130)(sin[tex]\pi[/tex]/3)([tex]\pi[/tex])(dr)^2
and then you take the integral to get
130*sin[tex]\pi[/tex]/3*[tex]\pi[/tex][tex]\int[/tex]dr^2
and I'm not sure when to go from here...
 
Last edited:
  • #4
So I was pretty sure I had it when I realized you can just take the dot product between the area and the electric field.

Therefore I was led to this answer.
A dot E= 130 * pi(.05)^2 * cos 66
Which is equal to the magnitude of the electric field multiplied by the magnitude of the area multiplied by the angle in between the two vectors (reference the pic above). However, my online homework program quickly told me I was wrong! Blergh! Any suggestions?
 
  • #5
Hi cwatki14! :smile:

(have a pi: π and a dot: · and try using the X2 tag just above the Reply box :wink:)
cwatki14 said:
So I was pretty sure I had it when I realized you can just take the dot product between the area and the electric field.

That's right … you only need to integrate if the field isn't constant, or if the surface (or its boundary) isn't planar. :smile:
Therefore I was led to this answer.
A dot E= 130 * pi(.05)^2 * cos 66
Which is equal to the magnitude of the electric field multiplied by the magnitude of the area multiplied by the angle in between the two vectors (reference the pic above). However, my online homework program quickly told me I was wrong! Blergh! Any suggestions?

Looks ok to me :confused:

(Are you sure it isn't 150, not 130?)
 

1. What is flux and why is it important?

Flux is the measure of the flow of a vector quantity through a surface. In the context of electric fields, it represents the number of electric field lines passing through a given area. It is important because it helps us understand the strength and direction of the electric field at a particular point in space.

2. How is flux calculated for an electric field?

To calculate the flux of an electric field through a surface, we use the formula: Flux = E * A * cosθ, where E is the magnitude of the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface normal.

3. What is the significance of the circle in this problem?

The circle represents the surface through which we are calculating the flux. In this case, the electric field is passing through the circle at a perpendicular angle, making the angle between the electric field and the surface normal 90 degrees.

4. How does the direction of the electric field affect the flux?

The direction of the electric field determines the angle θ in the flux calculation. If the electric field is perpendicular to the surface, the flux will be maximized. If the electric field is parallel to the surface, the flux will be zero.

5. Can flux be negative?

Yes, flux can be negative. This occurs when the electric field and the surface normal are in opposite directions, resulting in a negative angle θ. Negative flux indicates that the electric field is directed away from the surface.

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