# Flux Through a Non-Concentric Sphere

• mmdts

## Homework Statement

Assume I want to calculate the electric flux through a spherical surface centred at point P with radius R which contains a point charge Q, that is not concentric with the spherical surface.

Here, I can no longer assume that ∫∫sEdA = E.A, and I have to calculate the value of the surface integral to reach a result where the flux is a function of position on the sphere.

However, since I did not study surface integration yet, I decided to research it, when I got stuck in the calculations (or misunderstood something).

I want to know how to get the flux as a function in this case.

## Homework Equations

Gauss Law
Coulombs Law

@collinsmark

I can no longer assume that ∫∫sEdA = E.A,
You appear to be saying you could have used that formula had it been concentric, but strictly speaking that would not be the case. The right hand side would not be a dot product of two vectors, since the directions of the neither the field nor the surface vector are constant over the surface. But since they are everywhere parallel it would be correct to write |E| |A|.
I have to calculate the value of the surface integral to reach a result where the flux is a function of position on the sphere.
Why do you need to perform an integration? The flux as a function of location is a local value. Yes, you would need to integrate it if you then wanted to deduce the total flux (though obviously this would be unnecessarily complicated as a way to determine that).

Hello mmdts,

Allow me to add something to this. Gauss' Law can only be used in practice to find the strength of the electric field on a surface, if the surface is symmetrical with the charge. In other words, you can use Gauss' Law to find the unknown $\vec E$ only if the charge distribution is of one of the following:
• is spherically symmetric (points, spheres, etc.),
• is an infinite line charge (cylindrically symmetric, e.g., infinite lines, infinite cylinders, etc.), or
• is infinite plane charge (or infinite slab).
And that's pretty much it. And that's because those are the only cases where the magnitude of the electric field at the surface, $| \vec E | = E$, is a constant.

But Gauss' Law,
$$\oint_S \vec E \cdot \vec{dA}= \frac{Q_{enc}}{\varepsilon_0}$$
still holds true regardless of whether there is symmetry or not. :)

If you had an expression for $\vec E$ for all space, and had a formula for an arbitrary shape/strangely shaped closed surface, you could find the charge inclosed within the surface, even if there was no symmetry at all. Sure, you may not be able to use Gauss'Law to find $\vec E$ in the case where you didn't already know it, but Guass' Law still holds true anyway. And if you did have an expression for $\vec E$ you could use Gauss' Law to find the charge within.

This may be important here because in your original post you mentioned you were trying to find the flux $\Phi_E$, not the field E.

The flux through a flat, open (yet finite area A) surface is $\Phi_E = \vec E \cdot \vec A = EA \cos \theta$. To visualize that, think of "electric field lines" passing through the surface. The flux is proportional to the number of electric field lines that pass through*.

You can do the same visualization with a closed surface and a charge somewhere inside. The total flux through the closed surface is [proportional to] the total number of field lines that "exit" of the surface, minus the number of field lines that "enter" the surface. (In other words, the flux through that surface is proportional to the net number of electric field lines leaving the surface.)

Knowing what you know about electric fields being divergent, what does that tell you about the electric field lines leaving a spherical surface, even if you move the charge away from the center?

*(This analogy of "electric field lines" has its shortcomings. I don't mean to imply that electric fields really have a finite number of discrete lines. It's just an analogy to aid visualization.)

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Allow me to follow up with an exercise for you. If you can successfully complete this exercise, you should have a firm grasp of Gauss' Law, flux, and what it all really means.

In Figure 1a I drew a point charge surrounded by a spherical, Gaussian surface. I actually drew it as a two-dimensional circle, but you can think of it as a three dimensional sphere if you wish. The diagram shows eight "lines" of electric field (eight lines of flux, if you'd rather call them that). The number of field lines is proportional to the charge. I'll quantify the charge as 8 μC.

Now let's take each case, one at a time.

(a) As I mentioned in my last post, the total flux is proportional to the net number of field lines leaving the surface. So how many (net) lines of electric field (lines of flux) leave the surface? So what is the total flux? [Hint: Each line of electric field in this case corresponds to (1 μC)/ε0 of flux.]

(b) What if we change the size of the sphere? Recall that the surface area of a sphere is proportional to r2. Also recall that the electric field is proportional to 1/r2. Find the flux in this case. (Hint: So once again it all comes down to the net number of electric field lines/flux lines leaving the surface.)

(c) What if the charge is no longer in the center of the sphere? Calculate the number of lines leaving the surface and calculate the corresponding total flux.

(d) Now calculate the total (net) flux if instead of a spherical surface, we use an arbitrarily shaped, closed surface. Don't forget that if a line leaves the surface it counts as a positive amount of flux, and if it enters the surface it counts as a negative. A given line may leave and/or enter the surface more than once. Your goal is to find the net number of lines exiting the surface, and the corresponding total, net flux.

(e) What if we move the charge outside the sphere? Find the total amount of (net) flux leaving the surface in this case.

(f) Same as above except with an arbitrarily shaped, closed surface. Figure 1.

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The flux for a, b, c, d is the same, and the flux for e and f is zero. Thank you for helping me understand this, even though I just read your reply over 2 months late. After I read the reply, I did a little more reading and I think I understand it more clearly now.