Flux through concentric spheres

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SUMMARY

The discussion centers on calculating the total electric flux through a concentric spherical surface with a radius of 4 cm, given a uniform surface charge density of 4.0 nC/m² on an inner sphere with a radius of 2.0 cm. The correct approach involves first determining the total charge on the inner sphere by multiplying the surface charge density by the surface area, then applying Gauss' Law to find the electric flux. The final answer for the electric flux is 2.3 N·m²/C, confirming that the initial method of directly calculating the electric field was incorrect due to misunderstanding the charge distribution.

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  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric field calculations using Coulomb's Law
  • Knowledge of surface charge density and its implications
  • Basic proficiency in manipulating equations involving ε₀ (epsilon naught)
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  • Study the application of Gauss' Law in various geometries
  • Learn how to calculate total charge from surface charge density
  • Explore electric field concepts around spherical charge distributions
  • Review the implications of charge distribution on electric flux calculations
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Students and professionals in physics, particularly those studying electrostatics, as well as educators looking for clear examples of applying Gauss' Law and calculating electric flux.

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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



\Phi = E 4 \pi r^2 = \frac{q}{\epsilon_0}


E=k_e \frac{Q}{r^2}

The Attempt at a Solution



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

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

then used this to find the flux:

22250 \times 4 \times \pi \times (4 \times 10^{-2})^2 =447.36

But what's wrong with my approach? This is very different from the correct answer (2.3 N.m2/C).
 
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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 \epsilon_0 as per your stated formula for \Phi.
 
gneill said:
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 \epsilon_0 as per your stated formula for \Phi.

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

Like how do you know it is the green one and not the purple one?
 
Last edited by a moderator:
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 q/\epsilon_0.
 

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