anonymous
  • anonymous
A point charge produces an electric flux of +516 N* m2/ C through a gaussian sphere of radius 17.6 cm centered on the charge. What is the flux through a gaussian sphere with a radius 30.1 cm? What is the magnitude of the charge?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Kainui
  • Kainui
|dw:1350165391666:dw| Now imagine or draw out a smaller gaussian sphere with the same charge in it. What would the flux look like then? A good thing to understand is, what the hell is flux? Flux is the amount of flow through something else, and it's really that simple. Let's look at the confusing looking formula and explain it so we can understand. For this example, instead of looking at a whole sphere, we'll just look at one square. \[\phi _{E} =\int\limits_{}^{}E*dA\] |dw:1350165675568:dw|
Kainui
  • Kainui
Now we can see that the field lines are perpendicular to the area, so this is when we'll have maximum flux. If you flip the area block or the direction the field lines are coming from to look like this: |dw:1350165855493:dw| You see how there's absolutely no lines going through? This is 0 flux. So how do we tell the difference, well since that's the "dot product" that just means cosine of the angle. Flux=EAcos(theta) So when the angle is 90 degrees like in the first example, cos(90)=1 and flux is just E*A When the angle is 0 degrees like the second example, cos(0)=0 and there is no flux. So back to your spheres. All the electric field lines are going through the sphere at every point, so your flux is equal to the surface area of the sphere times the electric field going through it which is proportional to your charged particle! I can clarify anything if you need me to!
anonymous
  • anonymous
so its 4pir^2 * KeF/r^2 ?

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Kainui
  • Kainui
Almost, except E=Keq/r^2. Interesting thing about this is that it divides out r^2 and you can see that flux is truly independent of the size of anything it encloses.
anonymous
  • anonymous
so is now keq * 4pi?
Kainui
  • Kainui
Exactly, and this is an entirely general equation for things of any shape. So if you had a charge contained within a jar, a pot, a ball of clay, or your dishwasher, the flux would just be ke*4pi*q. Another note to be made is that the coulomb constant is actually discovered to be: \[Ke=\frac{ 1 }{ 4\pi \epsilon _{0} }\] So if you substitute in this, the 4pi divides out. Epsilon is the permitivity of free space and there's a corresponding one for magnetism called the permeability of free space with the symbol mu. Don't worry about them, but just know that hey, these things exist and you'll learn about them later and are used to calculate the speed of light which is an electromagnetic wave!
anonymous
  • anonymous
so now q/Eo?
Kainui
  • Kainui
Exactly.
anonymous
  • anonymous
sorry if i may ask,whats Eo?
Kainui
  • Kainui
That's epsilon, a greek letter that looks like E. I describe it above.
anonymous
  • anonymous
does it have a value?
anonymous
  • anonymous
i mean my answer becomes 516/? =
Kainui
  • Kainui
http://en.wikipedia.org/wiki/Vacuum_permittivity ε0 ≈ 8.854187817620... × 10−12
Kainui
  • Kainui
that's *10^-12
anonymous
  • anonymous
not accepting my answer :(
anonymous
  • anonymous
516/(8.85*10^-12) = 5.827*10^13N*m^2/C
anonymous
  • anonymous
is it the right unit?

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