# Electric Field; quick question

## Homework Statement

A charged particle is held at the center of two concentric conducting spherical shells. Figure a shows a cross section. Figure b gives the net flux Φ through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere. The scale of the vertical axis is set by Φs = 2.4 × 105 N•m2/C. What is the net charge (in Coulombs) of shell B?

heres the picture:
http://i36.photobucket.com/albums/e47/jo860/q18.jpg

## The Attempt at a Solution

Do i take the flux from each section and add them up?
Because in the tutor, it said that the net flux for a gaussian shell with radius r in between the radius of shell a and b was just the flux in between them and not the flux between them + the flux inside shell a
so im a little confused...
becaus i thot net flux was ALL the flux inside that radius r summed up

Last edited:

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collinsmark
Homework Helper
Gold Member
Perhaps I can help by just discussing Guass' law, and what it means.

Guass' law.
$$\oint_S \vec E \cdot d \vec A = \Phi_{electric} = \frac{Q_{enc}}{\epsilon_0}$$

Let's start with just discussing what electric flux is. Electric flux is the the number of electric field lines passing through some area. Keeping general (using a surface that is not necessarily closed),

$$\Phi_{electric} = \int_S \vec E \cdot d \vec A$$

But you need to have some sort of area to have flux. Imagine you have a small hoop that you could fit in the palm of your hand, and you bring it near an electric charge. The flux through that hoop is the total number of electric field lines passing through that hoop. If you bring the hoop closer to the charge, the flux increases. If you keep the hoop at a constant distance from the charge, and increase the hoop's radius, the flux increases. You can also change the amount of electric field lines passing through the hoop by changing the hoop's orientation. So,
.....(a) If the electric field intensity increases, the flux through the hoop increases.
.....(b) If the area of the hoop increases, the flux through the hoop increases.
.....(c) The more parallel the hoop's normal is to the electric field lines, the greater the flux.
And keep in mind, there might be electric field lines throughout all of 3D space. But concerning the flux, the only thing that matters is the surface in question (in this case, the surface defined by the hoop). If electric field lines don't pass through the hoop, they don't matter and don't effect the flux through the hoop.

Now let's move on to a closed surface. Instead of a hoop, let's enclose the charge in a hypothetical, spherical shell (not a real shell -- just use your mind's eye to make the shell). This hypothetical shell is called a Gaussian surface. The sum of all the electric field lines exiting this surface (from the inside out), minus all the electric field lines entering this surface (from the outside in), is the total flux.

Suppose we have a charge in this Guassian surface, and vary the radius. Suppose we increase the radius of this hypothetical, spherical shell (even though the amount of charge inside remains the same). The area increases, so the flux should increase too, right? No. As the radius increases, the electric field strength decreases proportionally so that the total flux remains constant (as long as the amount of charge within the surface remains constant). Another way to look at it is if you add up all the electric field lines exiting the surface minus the lines entering the surface, the size of the surface makes no difference (one again, assuming that the total amount of charge inside the surface remains constant). This is the essence of Gauss' law.

Okay, what if there is no charge inside the Guassian surface, but there is a charge just outside of it? Well, every electric field line that enters the surface will also leave the surface at some other point. So the total lines leaving minus the total lines entering equals zero. So in conclusion:

Gauss's law states that the net number of electric field lines exiting the closed, hypothetical, Gaussian surface, is proportional to the charge within that surface (charges outside the surface don't matter). And the constant of proportionality is 1/ε0.

So use Guass' law to find the total charge within the surface. Then, if you happen to know that there are two charges within that surface, and you know the value of one of them, the other must be....?

And if you calculate the total charge within a Gaussian surface at a different radius, and you know that there are three charges within that surface, and you know the value of two of them, the third must be....?

• sparkie