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Gauss's Law and Electric fields

  1. Mar 5, 2013 #1
    It is asked to find the electric flux through a "gaussian" sphere which has a point charge (-3 microcoloumbs) enclosed with the radius of 0.2 m ... I can shortcut this and use Flux= q Enclosed / e0.. However, i want the other approach using the formula Flux = ∫ E da... I know that E is constant at the surface but shouldn't we be adding them to get amuch larger E... I mean E is constant in magnitude for every point on the sphere. Shouldn't we add them up. A much general rule to discuss is that why do we sometime need double integration in such formulas. I mean consider a cube. Even if E is changing, it will be constant at any face...
  2. jcsd
  3. Mar 5, 2013 #2

    Doc Al

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    You are taking each element of the surface area and multiplying it by the value of the field at that point to get E*da (the scalar product). You are adding those up when you integrate. Assuming that the charge is in the center of your sphere, the integration is easy, since E is a constant over the surface of the sphere. If E is not constant over the surface, you'll have a harder time evaluating the integral.
    I'm not sure what you are asking here. If you used a cube as your surface instead of a sphere, the value of E would vary over the face of the cube. (Giving you a messier integral.)
  4. Mar 5, 2013 #3
    A point charge will give always a spherically symmetric field. Therefore the easiest thing to do is to integrate on a sphere with the charge in the center. This because in this way you will be sure that the electric field will be constant and radially directed, and therefore (i) the scalar product with the surface element will be trivially the product of the moduli (without factors from the angles as ##\cos 0=1##); moreover (ii) you will be able to take out the field from the integral.

    This is the only way to use this theorem if you want to find the expression for ##E(r)##, because in this way, taking out the field from the integral, you will have
    $$ E(r)=\frac{q}{\int \mathrm{d}A}=\frac{q}{4\pi r^2} $$

    If you integrate on something different from the sphere you cannot take ##E## outside the integral and therefore you can't find anything, because you would need ##E(r)## to be able to solve the integral, but this is exactly what you want to find and therefore you are stuck.
  5. Mar 5, 2013 #4
    Why would it be constant at a face? You would need a special configuration of charges for that to be true.

    Also, Gauss's Law is the general method. You just need to choose your coordinate system appropriately for the charge distribution of interest. In general, you are calculating a double integral. In the example of the sphere it is an integral over [itex]\theta[/itex] and [itex]\phi[/itex] for a constant value of radius. (The [itex]\phi[/itex] integral is often trivially [itex]2\pi[/itex].)
  6. Mar 6, 2013 #5
    The charge doesn't need to be at the center. E will always be constant over the surface of the sphere no matter where exactly the charge is as long as it is inside the sphere. Strictly speaking however this is only true as long as everything is static. If the charge inside moves around it induces a current in the sphere which makes the situation messier.
  7. Mar 6, 2013 #6


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    If the sphere is a physical spherical shell made of conducting material, then the potential V will be the same everywhere on the sphere, but the electric field E will be the same everywhere on the sphere only if the charge is at the center.

    If the sphere is a physical spherical shell made of non-conducting material, then both V and E will be the same everywhere on it only if the charge is at the center. (assuming that's the only "extra" charge involved, of course)

    However, in this thread, we're taking about a fictitious Gaussian surface, whose only purpose is to give us something to integrate E over.
  8. Mar 6, 2013 #7
    This is not true. You can easily see that by evaluating the field at points on the sphere. The magnitude of the separation vector is no longer the same at each point on the sphere, so the field cannot be the same at all points on the sphere. You would be correct in saying that the net flux is still proportional to the enclosed charge, though.
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