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Net Flux through a sphere (Gauss's Law)

  • Thread starter Alex G
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Homework Statement



Find the net electric flux through the spherical closed surface shown in the figure below. The two charges on the right are inside the spherical surface. (Take q1 = +1.96 nC, q2 = +1.03 nC, and q3 = -3.09 nC.)

https://www.webassign.net/serpse8/24-p-006-alt.gif


The Attempt at a Solution



I know this is a Gauss's Law problem. I began by treating each of the charges INSIDE the sphere as a point charge form Coulomb's Law. I am under the impression that the one outside has no effect on the flux through the sphere?
Thus I set up the equations
k(q1/r^2) + k(q2/r^2) = Net Flux
However, I do not know the radius of the sphere. So I assume this is not the right way to go about this.
I am also aware that the Electric field is radial.
 

Answers and Replies

  • #2
Doc Al
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I know this is a Gauss's Law problem.
What does Gauss's law state about the flux through a closed surface?
I began by treating each of the charges INSIDE the sphere as a point charge form Coulomb's Law.
Why? Stick to Gauss's law.
I am under the impression that the one outside has no effect on the flux through the sphere?
Right!
Thus I set up the equations
k(q1/r^2) + k(q2/r^2) = Net Flux
No, that's not the net flux.

It's much easier than that. :wink:
 
  • #3
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Ooooh, I love the break down (:

Lets see. Gauss's Law states for an enclosed surface the flux is the
Integral(electric field {dot product} the infinitesimal areas)

In this case, E is constant and the sum of the infinitesimal areas is 4(pi)r^2.

So, err, is this right so far? (: This is the first problem of this type I've done.
 
  • #4
Doc Al
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Lets see. Gauss's Law states for an enclosed surface the flux is the
Integral(electric field {dot product} the infinitesimal areas)
No, that's just the definition of the electric flux. What does Gauss's law say that the flux must equal?

In this case, E is constant and the sum of the infinitesimal areas is 4(pi)r^2.
FYI, E is not constant!
 
  • #5
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Oh, I guess it isn't equal, my mistake, I got ahead of myself when I saw a sphere and assumed the constant radius meant constant E, but they are not centered. (:

Ah, ha! Gauss's Law states that it must = qenclosed / Epsilon0

So, if Epsilon0 = (1/4(pi)k)

Then is it: (the sum of the enclosed charges) over (1/4(pi)k)?

Sorry, I think I'm starting to get it, or just narrowing down everything and getting rid of over thinking! I'm using my notes and he isn't the best at teaching /-:
 
  • #6
Doc Al
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Ah, ha! Gauss's Law states that it must = qenclosed / Epsilon0

So, if Epsilon0 = (1/4(pi)k)

Then is it: (the sum of the enclosed charges) over (1/4(pi)k)?
That's it. Much easier than trying to calculate the flux directly, eh?
 
  • #7
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Yay! Thank you! (:
Since q2 and q3 are the only affecting charges:

(1.03*10-9-3.09*10-9)/ (1/4*(pi)*8.9876*109)?

That appears to come out to -2.9183246*10-19, does that seem reasonable for fluxes? My teacher hasn't used any numbers in class /-:
 
  • #8
Doc Al
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Since q2 and q3 are the only affecting charges:

(1.03*10-9-3.09*10-9)/ (1/4*(pi)*8.9876*109)?
This looks OK. Realize that you are dividing by ε0 = 1/(4πk). (It might be easier for you just to look up the value of ε0.)

That appears to come out to -2.9183246*10-19, does that seem reasonable for fluxes?
Double check your arithmetic.
 
  • #9
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Haha! YES! Thank you! And it all seemed to click much better than before!

(Yes using Epsilon0 was the best way to go)

Answer ended up at -232.658 N*m2/Cwhich makes much more sense.

So I would probably use the integral definition of Gauss's Law when given an object with no value of the charges? Or rather given the value of the Electric field?
 
  • #10
Doc Al
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So I would probably use the integral definition of Gauss's Law when given an object with no value of the charges? Or rather given the value of the Electric field?
If you're given the field everywhere along a closed surface, you can use Gauss's law to figure out the net charge enclosed. But then you'd have to calculate the flux directly. Sometimes that's easy; often it's not.
 

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