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PF Gold
P: 3,188
 Quote by sunjin09 The δ(θ-pi/2) part of your expression is missing a 1/r factor, if you think about the difference between dθ and dz. What you want is an infinitesimal displacement in the z direction, i.e., dz=r*dθ at pi/2. BTW in cylindrical system δ(r,r')=$\delta(r-r')\frac{\delta(\phi-\phi')}{r}\delta(z-z')$, where the 1/r is associated with the delta of $\phi$ coordinate, for the same consideration, but z coordinate is unaffected.
What I don't understand is why do I have to use "δ(r,r')"?
For instance in cylindrical coordinates, I made the ansatz that the solution to the problem is $\rho (\vec x)=C \Theta (r-R) \delta (z)$. When integrated over all the space, this gave me $C=\frac{Q}{\pi R^2}$ leading to $\rho (r,z)=\frac{Q}{\pi R^2}\Theta (r-R) \delta (z)$ which coincides with the answer found on the Internet. I tried the exact same method in spherical coordinates but for some unknown reason to me, the answer found on the Internet differs from mine.
In either cases, I didn't use the expression for δ(r,r').