# Homework Help: Which equation for flux?

1. Dec 18, 2011

### ZedCar

Calculate the flux, due to a point charge of 100 nC at the origin, that penetrates a
sphere of radius 0.01 m centred at the origin.

Would anyone know which equation I should be using for this problem?

Thank you.

2. Dec 18, 2011

### cepheid

Staff Emeritus
Physics is not about blindly choosing equations and plugging numbers into them. It is about understanding physical concepts (which are expressed mathematically). In this case, the concept you need to understand is, "what is the flux of an electric field across a surface?" Once you know the definition of electric flux, then you know everything you need to know to solve this problem.

3. Dec 18, 2011

### ZedCar

Well, as I understand it, electric flux is the flux of an electric field.

The electric field through a planar area has one means of evaluation. Since this question relates to a sphere (non-planar) then this evaluation would not apply.

The other evaluation, which applies when the area is non-planar, requires an integral.

So, the equation I was thinking of using is:

∅ = ∫EcosθdA

4. Dec 18, 2011

### cepheid

Staff Emeritus
Right, but that definition is kind of tautological and useless unless if you say what flux is. The flux of a vector field across a surface of some area is defined as the surface integral of that vector field over that area. In other words, it is the integral over the area of the electric field vector dotted with the unit normal vector at each point.

Well, actually, the definition in terms of the integral is the general defintion: it always applies to every case, including the planar one. It's just that if the surface is planar and the electric field is normal to it at every point (and uniform in magnitude everywhere) then the integral reduces to a simple multiplication.

This will work -- especially if you use spherical coordinates. Now, what is cosθ at every point on the sphere? Hint: the unit normal vector always points "radially outward" at every point on the surface of a sphere, and the electric field also happens to point "radially outward" everywhere.

This equation will work, but there is a spherical symmetry to the problem that you can exploit. (The properties of the electric field allow you to do so). If you exploit this symmetry, you won't have to do any integration at all. Have you heard of something called Gauss' Law? Using Gauss' Law, this is a 2-second problem with no math.

5. Dec 18, 2011

### ZedCar

= 1

The "other" equation;

∅ = EAcosθ
would then be
flux ∅ = (100x10^-9) x (4∏r^2)
= 1.256 x 10^-10 (Nm^2)/C

6. Dec 18, 2011

### cepheid

Staff Emeritus
Yeah. Did you try the integral, given that information?

Nope. This equation is what the integral reduces to under simpler circumstances. It is not applicable here and it is not Gauss' Law. Gauss' Law says that the flux of the electric field through any closed surface is just proportional to the net charge enclosed by that surface (hence no math required). I'll let you look up Gauss' Law to figure out what the constant of proportionality is. Or, you can just evaluate your integral to find out (i.e. derive Gauss' law yourself).

7. Dec 29, 2011

### ZedCar

So the answer is simply Q/ε0 ?

In this case, the radius is of no relevance in the calculation of the flux?

8. Dec 30, 2011

### cepheid

Staff Emeritus
Exactly. That is Gauss' Law.

There is no dependence on radius. If you had done the integral, you would have seen exactly how the r-dependence cancels out. But the basic idea is that the surface area increases as r2, which tends to increase the flux. However, the E-field strength decreases as 1/r2, which tends to reduce the flux. These two effects precisely cancel each other out, leading to no dependence on distance.