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 $q/\epsilon_0$.

 

[rwisz]

Your approach is incorrect because you are using the electric field due to the charge at the surface of the concentric spherical surface instead of the electric field due to the charge on the spherical surface itself. To find the total electric flux, you need to integrate the electric field over the entire surface of the concentric sphere. This can be done using the equation \Phi = \int \vec{E} \cdot \vec{dA}, where \vec{E} is the electric field and \vec{dA} is an infinitesimal element of area on the surface.

To find the electric field due to the charge on the spherical surface, you can use the equation E = k_e \frac{Q}{r^2}, where Q is the total charge on the spherical surface. In this case, Q = (4 \times 10^{-9} C/m^2) \times (4 \pi (0.02 m)^2) = 1.01 \times 10^{-6} C.

Using this, we can find the electric field at a distance of 4 cm from the center of the spherical surface:

E = (8.99 \times 10^9 N.m^2/C^2) \frac{1.01 \times 10^{-6} C}{(0.04 m)^2} = 2.25 \times 10^5 N/C

Now, we can use this electric field to find the total electric flux through the concentric spherical surface:

\Phi = \int \vec{E} \cdot \vec{dA} = \int_0^{2\pi} \int_0^\pi (2.25 \times 10^5 N/C) (4 \pi r^2 \sin\theta) d\theta d\phi = (2.25 \times 10^5 N/C) (4 \pi (0.04 m)^2) = 2.28 N.m^2/C

This is closer to the correct answer of 2.3 N.m^2/C. Please note that there may be small variations due to rounding errors.