Why is the Dirac Delta Function in the Charge Density Solution?

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The discussion centers on the presence of the Dirac delta function in the charge density solution derived from the electric potential V(r) = (Ae^(-pr))/(r). Participants explore the calculation of the electric field E and charge density ρ using the equations -∇V = E and ∇·E = ρ/ε₀. The inclusion of the Dirac delta function arises from the singularity at r = 0, indicating a point charge at the origin, which leads to a charge density that cannot be simplified by canceling terms without accounting for this singularity. It is emphasized that if the charge distribution exhibits a singularity, the Dirac delta function must be included to accurately represent the charge density. Understanding this concept is crucial for correctly solving problems involving point charges in electrodynamics.
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Homework Statement


the electric potential of some configuration of charge is given by the expression:

V(r)= (Ae^(-pr))/(r)

where A and p are constants. Find the E-field E(r), the charge density rho(r) and total charge Q. The answer for rho is given in book as:
epsilon-not * A * ((1/r)4* pi* dirac delta^3 (r) - p^(2) e^(-p*r))
(The stuff inside parentheses is all divided by r


Homework Equations


-del V= E; del dot E = rho/epsilon-not ;



The Attempt at a Solution


I the took gradient of V using sperical coordinates using the formula
-del V= E

to obtain E.

Then I used the formula

del dot E = rho/epsilon-not

to find rho. I am not getting the same rho as in the book. I don't understand why the dirac delta function is in the answer. Do you have to perform an integration or am I taking a wrong apprach all together?
 
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Can you show how you got your volume charge density? What equation did you strart from?
 
the formula for obtaing rho goes thru this eqn:

del dot E = rho/epsilon-not

Del dot E in sperical coords. is : (1/r^2) d/dr [ r^2 V(r)] (the (1/r^2) and r^2 are built into the formula.) once i did this multipled both sides by epsioln-not to obtain rho. This question is from Grifiths electrodynamics book ed. 5 problem 2.46
 
Then what did you do? When you took the derivatives, did you say

\frac{r^2}{r^2}\frac{dV}{dr}=\frac{dV}{dr}\;?

How justified are you in canceling out the r2 terms at r = 0? That's where the Dirac delta function comes in.
 
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This is what I have so far if you click on the link. Can I cancel out the r^2 terms in the differential? Or do I distribute the r^2 in the numerator and then perform the differential? How can I insert a dirac delta function? If I cancel out the r^2 in the differential I don't get the same answer as the book.
 
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If the charge distribution has no singularity (does not blow up) as r goes to zero, then you can cancel the r2 in the numerator and denominator and proceed merrily on your way. If there is a singularity, then you essentially have "zero divided by zero". You know that in this case there is a singularity because the electric field goes to infinity as r goes to zero. This means a point charge at the origin which requires a Dirac delta function in the volume charge density.
 
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