Flux through concentric spheres

[roam]

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).

 

[gneill]

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 $\Phi$.

 

[flyingpig]

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?

 

[gneill]

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 $q/\epsilon_0$.