# Homework Help: Need a physical picture

1. Oct 4, 2011

### humo90

Need a physical picture!!!

The electric potential of some charge configuration is given by V(r) = A*e^(-λr) / r

Its Electric field is E = A*[(λr+1)/(r^2)]*e^(-λr) $\hat{r}$

And its charge density is ρ = 4*pi*A*ε0*δ(r) - [(ε0*(λ^2)*A*e^-λr) / r] , where δ(r) is delta function.

The thing that surprised me is the total charge is zero (Q = 0), how could this happened since I have a charge density? and can someone help me answering the following two questions.

1. What is the physical picture of the charge density (=how does the charge density look
like)?

2. Where could a charge density of this form appear in nature?

2. Oct 4, 2011

### gash789

Re: Need a physical picture!!!

Hi, How have you figured out that the total charge is zero exactly?

For a picture of the charge density its worth simplifying the statement of charge density by putting all the constants together, unless I have made a mistake does this make sense?

$ρ(r)=B*δ(r)-C\frac{e^{-Dr}}{r}$

It is then a case of taking several significant values of radial points (it should already be clear it is spherically symmetric) and considering the value of the charge density given that B and C are positive. Does this help?

Once you have the shape and given the spherical nature the second question should make more sense.

3. Oct 4, 2011

### Dickfore

Re: Need a physical picture!!!

A charge distribution of this nature appears when you place a point charge in a plasma. The point charge is screened by the ions/electrons of the plasma.

4. Oct 4, 2011

### humo90

Re: Need a physical picture!!!

To get the total charge, I took the volume integral of the charge density, that is

Q = ∫ρ dv = ∫(4*pi*A*ε0*δ(r) - [(ε0*(λ^2)*A*e^-λr) / r]) dv

= (4*pi*A*ε0) ∫δ(r) dv - (ε0*(λ^2)*A) ∫((e^-λr) / r) dv

= (4*pi*A*ε0) - (ε0*(λ^2)*A) ∫∫∫((e^-λr) / r) (r^2)*sin(∅)*dr*d∅*dθ,

since ∫δ(r) dv = 1

and dv = (r^2)*sin(∅)*dr*d∅*dθ in spherical coordinates.

Also, ∫∫∫((e^-λr) / r) (r^2)*sin(∅)*dr*d∅*dθ = (4*pi*ε0*(λ^2)*A)((1/λ^2)-((r*e^-λr)/λ)-((e^-λr)/(λ^2)),

and for entire space: r goes to ∞, we get ∫∫∫((e^-λr) / r) (r^2)*sin(∅)*dr*d∅*dθ = (4*pi*ε0*A).

Hence, Q = (4*pi*A*ε0) - (4*pi*ε0*A) = 0

I am really still confusing about understanding this problem.

5. Oct 4, 2011

### gash789

Re: Need a physical picture!!!

Hi, your working appears correct. I am not 100% but taking Dickfore answer this solution would make sense. The total enclosed charge is zero due to screening by the plasma. Sorry I cant be of any help

6. Oct 4, 2011

### ehild

Re: Need a physical picture!!!

There is a positive point charge in the middle and negative charge distribution around. It is somewhat similar to a spherical capacitor with a very small inner sphere, with positive charge, and a grounded outer sphere where negative charges accumulate. There is an electric field between the spheres, there is a charge distribution, and the total charge of the capacitor is zero.

ehild