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"the electrostatic field of a point charge q is, where er is the radial unit vector. Find the divergence of E away from the origin, which is the point source of the field."
I've never taken the divergence of a field with only one component (r), so it seems too simple, but is it:
?
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Originally Posted by Roobydo
"the electrostatic field of a point charge q is
I've never taken the divergence of a field with only one component (r), so it seems too simple, but is it:
Try calculating the divergence now.
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ok,
?
They can't both be true.
My main question though was: since the field is only dependent on r, and the"vector" and E vector have only one component, is the divergence just dE/dr?
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This is not going well, you are forgetting the presence ofok,
They can't both be true..
You need to consider two things:
and
No.My main question though was: since the field is only dependent on r, and the
Last edited by macaw; 2011-May-23 at 12:38 AM.
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Please don't patronize me, I take this very seriously. I'm not a hobbyist, I'm a university student.
In the answer? Divergence is scalar.you are forgetting the presence of
Yes, but I would prefer not to change everything to cartesian coordinates; see below. I would also prefer to work with the unitized radial vector.You need to consider two things:
and
I see that now that I have three components. But intuition tells me that I shouldn't need three: I know the field is spherically symmetrical about the origin, so the divergence should be a function of radius alone.
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No, it isn't. If you insist on working in polar coordinates, then you need to remember that:"the electrostatic field of a point charge q is
I've never taken the divergence of a field with only one component (r), so it seems too simple, but is it:
where
You should get the same result , whether you are working in polar or cartesian coordinates.
Last edited by macaw; 2011-May-23 at 03:47 PM.
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if you do your calculations correctly , you'll find out that the result does not depend on the radius at all. You should get the same result, independent of the system of coordinates. This should help you in confirming that your calculations are correct.
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If you're going to give me equivalences to remember, I ask that you tell me how you arrived at them.No, it isn't. If you insist on working in polar coordinates, then you need to remember that:
where
r2Er is a constant, so what you're saying is
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Yes. If you work the problem in cartesian coordinates you should get the same answer.If you're going to give me equivalences to remember, I ask that you tell me how you arrived at them.
r2Er is a constant, so what you're saying is
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What about Gauss' law:
Where doescome from?
Order of Kilopi
Gauss' law is a good way to see that the answer must be zero everywhere away from the charge, since rho=0 everywhere but the origin. Indeed, Gauss' law is the statement that charges are what create a local divergence of the field, so surround the charges with a "Gaussian pillbox" that conforms to the symmetry of the charge distribution to get the field that threads the surface of the pillbox. But that's how you get the field, which is what you usually want to know-- this problem asks you to demonstrate that you get that field by integrating a divergence that here is nonzero only at the origin. You can also see that by using two spheres of different radius, both centered at the origin. Every field line that threads the inner sphere also passes out of the larger sphere, because the density of field lines falls like 1/r^2, and the surface area rises like r^2, so the number of field lines stays constant. That means there is zero divergence of the field between the spheres. Since anywhere but the origin could be "between the spheres", everywhere but at the origin the divergence is zero. Yes, the divergence is infinite at the origin in this idealized charge distribution.
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From here, for example. If you study physics, you should have had some advanced calculus.
Where does
Last edited by macaw; 2011-May-23 at 03:48 PM.
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I study math- the HW is for a grad level applied math class- I've only taken a semester of physics (I hope to take more). A problem I keep coming across is that they tend to teach us the basics, and skip over the details to save time assuming we'll figure them out if we ever really need them. Alternate coordinate systems almost always fall to the wayside.
That site is a great resource too, thanks.
And thanks for the input Ken, it's always good to hear from you.
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If this is the case, then you should work out the problem in cartesian coordinates, the way I started you. Did you work it out in cartesian coordinates?
Yes, it is.That site is a great resource too, thanks.
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Double checking: you're specifying the partial r of spherical, and neglecting partial phi and partial theta, which do not change in this problem. ? The trick it seems (and it got me as I lurked) is to instantly know/recall there's no coordinate system of 'just r'. You have to go full spherical coordinates, which then allows Product Rule.
Do you need a pay account to post these equations from codecogs?
Order of Kilopi
See the F.A.Q.
[tex]E_r=\frac{q}{4\pi\epsilon_{0}r^2}[/tex] ... gives ...
Get up, a get-get, get down.
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Thanks!
Any nits macaw?Particularly carriage of vector from del to r-hat.
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Correct.
No.Do you need a pay account to post these equations from codecogs?
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