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

[richard7893]

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?

 

[kuruman]

Can you show how you got your volume charge density? What equation did you strart from?

 

[richard7893]

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

 

[kuruman]

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.

 

[richard7893]

http://s22.photobucket.com/albums/b317/richard7893/?action=view&current=emag.jpg"
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.

 

[kuruman]

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.