Flux through concentric spheres

  • Thread starter roam
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  • #1
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Homework Statement



Charge of uniform surface density (4.0 nC/m2) is distributed on a spherical surface (radius = 2.0 cm). What is the total electric flux through a concentric spherical surface with a radius of 4 cm?

Homework Equations



[tex]\Phi = E 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]


[tex]E=k_e \frac{Q}{r^2}[/tex]

The Attempt at a Solution



I tried finding the electric field due to the charge at the surface of the concentric spherical surface

[tex]E = (8.9 \times 10^9) \frac{(4 \times 10^{-9})}{(4 \times 10^{-2})^2} = 22250[/tex]

then used this to find the flux:

[tex]22250 \times 4 \times \pi \times (4 \times 10^{-2})^2 =447.36[/tex]

But what's wrong with my approach? This is very different from the correct answer (2.3 N.m2/C).
 

Answers and Replies

  • #2
gneill
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You're given the charge density on the inner sphere, not the total charge. You should first work out the total charge on the surface of the sphere, then divide by [itex]\epsilon_0[/itex] as per your stated formula for [itex]\Phi[/itex].
 
  • #3
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You're given the charge density on the inner sphere, not the total charge. You should first work out the total charge on the surface of the sphere, then divide by [itex]\epsilon_0[/itex] as per your stated formula for [itex]\Phi[/itex].

How do you know the charge density is in the inner surface and not inside the surface?

[PLAIN]http://img707.imageshack.us/img707/6093/unledqqn.jpg [Broken]

Like how do you know it is the green one and not the purple one?
 
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  • #4
gneill
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2,866
The usual interpretation of the phrase, "distributed on a spherical surface" is that it is located on the outer surface unless otherwise indicated.

Even so, it would not make a difference given Gauss' Law which proclaims that the flux through any closed surface surrounding a charge q is equal to [itex]q/\epsilon_0[/itex].
 

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