# Gauss' Law: Charge distribution on concentric spherical surfaces

1. Mar 12, 2017

### Arman777

1. The problem statement, all variables and given/known data

A metallic sphere of radius a is placed concentrically with a metallic spherical shell with inner radius b and outer radius c. The sphere has a total charge of 2Q and the shell has a total charge of 3Q.
(a) What is the charge distribution? Specifically, what is the charge density in the interior of the metals and at each surface (at r = a, b and c)?
(b) Use Gauss’ law to ﬁnd the electric ﬁeld at every point.

2. Relevant equations
Gaussian Law

3. The attempt at a solution
Part (a) The charge density of the interior of A is $ρ=\frac {3(2Q)} {4πa^3}$ and for other sphere
$ρ=\frac {3(3Q)} {4π(c^3-b^3}$
The surfaces charge density should be zero.I think like that cause.Electrons cannot pass air.
Electric field will be zero inside the small sphere $(E=0 , r<a)$ then its $E=\frac {2Q} {4πr^2}$ , $a<r<b$ then again zero inside the outer sphere $(E=0, b<r<c)$ then its $E=\frac {5Q} {4πr^2}$ , $c<r$

Is this true ?

2. Mar 12, 2017

### BvU

Nope. Charge in conductors is totally mobile. If the charge carries get a chance, they will sit as far away from each other as possible until they can't go any further. (Mathematically: $\Delta V = 0$ -- or rather $\vec \nabla V = 0$ -- in a conductor).

You only show answers, not your work; please show the steps and comment so we can find out whether
holds ...

3. Mar 12, 2017

### Arman777

If my answers are wrong my logic ahould be wrong too...The problem is Does the -2Q charge will cover the surface of inner sphere.I assumed no cause there no metal betweeen those two charges.But If they can still move then things will change for sure.

4. Mar 12, 2017

### ehild

-2Q is the amount of excess charge on the metal sphere, not two pieces of charge. Charge is not an object, it is attribute of objects. Just like an object have mass, it can have charge. If you charge a metal sphere for -2 C, it means lot and lot charge carriers, electrons. One electron has charge of -1.6˙ 10-19 C, so -2 C means 1.25˙10 19 electrons.˙

5. Mar 12, 2017

### Arman777

Sure but those -2Q charges will cover the surface ? I mean when I am trying to calculate the surface charge density I ll use that -2Q quantitiy

6. Mar 12, 2017

### ehild

Yes, they cover the surface. Electrons move free on the metal surfaces, so one electron can be anywhere there. You can calculate the surface charge density by dividing the charge with the surface area.

7. Mar 12, 2017

### Arman777

"what is the charge density in the interior of the metals and at each surface " Then the charge density is 0. but surface charge density is $E=\frac {-2Q} {4πr^2}$ ?

8. Mar 12, 2017

### BvU

Yes. Don't use the symbol $E$ (it is generally used for the electric field strength; the symbol $\sigma$ is most often used for surface charge density.) And make a habit of mentioning the units too.

9. Mar 12, 2017

### Arman777

Yeah my mistake sorry

10. Mar 12, 2017

### BvU

Note that there are three surfacess for which you are asked to give the charge density...

11. Mar 12, 2017

### Arman777

So the outer shell inner surgace will be +2Q and outer surface +Q ?

12. Mar 12, 2017

### BvU

Spot on ! Bingo ! However: that's the charges, not the charge densities but I suppose you understand what goes on.

13. Mar 12, 2017

### Arman777

I think I understand the idea...Electric field will be zero every point inside the outer shell till the outer surface then its 3Qk/r^2 ?

14. Mar 12, 2017

### Arman777

Yeah I understand Just have to divide surface areas.And outher shell surface charge density will be 3Q/(Volume of outer shell)

15. Mar 12, 2017

### BvU

Oops, wrong dimension. You want something in C/m2 !

16. Mar 12, 2017

### Arman777

There should be no surface...

17. Mar 12, 2017

### Arman777

I ll write proparly..everything

18. Mar 12, 2017

### Arman777

Lets call the spere inside A and outside sphere B.The charge density of A, $ρ=0$ (Total charge is zero)
Outer Surface charge density of A $σ=\frac {-2Q} {4πa^2} \frac{C} {m^2}$
İnner surface charge density of B $σ=\frac {2Q} {4πb^2} \frac{C} {m^2}$
Charge density of B $ρ=\frac {3(3Q)} {4π(c^3-b^3)} \frac{C} {m^3}$
Outer surface charge density of B $σ=\frac {Q} {4πc^2} \frac{C} {m^2}$

Electric Field will change like;
$E=0$ till outer surface of B.(till r=c) then it will be $E=\frac {3Q} {4πε_0r^2} \frac{N} {C}$

Electric field should be zero inside the shell B.

19. Mar 12, 2017

### BvU

Not good: check total charge on B

20. Mar 12, 2017

İts 3Q