# Flux through concentric spheres

## 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).

## Answers and Replies

gneill
Mentor
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 $\Phi$.

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 [Broken]

Like how do you know it is the green one and not the purple one?

Last edited by a moderator:
gneill
Mentor
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 $q/\epsilon_0$.