I took the divergence of the function 1/r^{2}[itex]\widehat{r}[/itex] in spherical coordinate system and immediately got the answer as zero, but when I do it in cartesian coordiantes I get the answer as 5/r^{3}. for [itex]\widehat{r}[/itex] I used (xi+yj+zk)/(x^{2}+y^{2}+z^{2})^{1/2} what am i missing?
Hi Idoubt! If I take the cartesian derivative with respect to x I get: [tex]{\partial \over \partial x}{x \over (x^2+y^2+z^2)^{3/2}} = {-2 x^2+y^2+z^2 \over (x^2+y^2+z^2)^{5/2}}[/tex] Of course the derivatives wrt to y and z have to be added yet... How did you get 5/r^{3}?
sorry I made a silly error in my calculation [itex]\frac{\partial}{\partial x}[/itex] x(x^{2}+y^{2}+z^{2})^{-[itex]\frac{3}{2}[/itex]} = (x^{2}+y^{2}+z^{2})^{-[itex]\frac{3}{2}[/itex]} - 3x^{2}(x^{2}+y^{2}+z^{2})^{-[itex]\frac{5}{2}[/itex]} When i add y & z parts it cancels and becomes zero, so the div of [itex]\frac{1}{r^2}[/itex][itex]\widehat{r}[/itex] is zero. The reason that confused me was Electric fields can have a non zero divergence. Is that because of the singularity at r=0?
Not every electric field is generated by exactly 1 symmetrical charge. And in practice there is no singularity, since the charge would have to be a point charge for that.
Isn't the field due to just one charged assumed to be a field created by a point charge? ( I mean it makes no difference if we make that assumption right? ) But, how can an electric field have a non zero divergence then?
Yes. It would only make a difference when you get very close to where the point charge is supposed to be. Take for instance 2 point charges and you'll have non-zero divergence.
But when we take two charges we no longer have the [itex]\frac{1}{r^2}[/itex] function ( unless we approximate at large distances) It is my understanding that even the electric field of a point charge has non zero divergence is it not so?
Right on both counts. :) My point about point charge is unrelated to divergence. Only that a real electric field has no singularity.
my question is , how can a field which is scalar multiple of the function [itex]\frac{1}{r^2}[/itex][itex]\hat{r}[/itex] have a divergence when i just proved that the divergence of this function is zero?
The divergence is not 0, it's [itex]4 \pi \delta^{(3)}(\vec{x})[/itex]. It's not too tricky to prove with help of Green's first integral theorem applied to the whole space with a little sphere around the origin taken out and then taking its radius to 0.
meh a and b are just constants and in this case F(X) is obviously the Function of X. The Greek symbol that looks like an O with a hat means Tensor. E(X) I would assume means the Energy of the system in question in terms of the vector.
If you've done matrices you'll know a tensor is expressed as a matrix of a matrix. A vector in this case is simply a matrix with a scalar. http://en.wikipedia.org/wiki/Vector_space So to steal more wiki imagery a tensor is:
hmm ok ok so thats a little advanced for me, but my question was about the divergence of the [itex]\frac{1}{r^2}[/itex][itex]\hat{r}[/itex] function. In simpler terms can you tell me the reason for non-zero divergence in the electric field of a point charge?
No, it's a distribution, the socalled Dirac [itex]\delta[/itex] distribution. It's a very useful concepts. It's even so useful that the mathematicians built a whole new subfield of mathematics to define this concept rigorously, functional analysis!
I suspect you're referring to Maxwell's law that says: [tex]\textrm{div }\vec E = \frac {\rho} {\epsilon_0}[/tex] The divergence of a spherically symmetric charge is zero everywhere except at the place where the charge actually is. A point charge is a special case, since the charge density would be infinite (actually a Dirac delta-function). In practice the charge would take in a certain volume. The divergence of the electric field is not zero at the place where the charge density is not zero.