# Homework Help: Potential vs Charge

1. Aug 10, 2016

### Fips

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

Two conducting spheres are far apart. The smaller sphere carries a total charge Q. The larger sphere has a radius that is twice that of the smaller and is neutral. After the two spheres are connected by a conducting wire, the charges on the smaller and larger spheres, respectively, are:
A. Q/2 and Q/2
B. Q/3 and 2Q/3
C. 2Q/3 and Q/3
D. zero and Q
E. 2Q and −Q

3. The attempt at a solution

I thought that charge would spread equally by the two spheres, but it seems that it deppends on the radius of them. Can someone tell me why? mathematically if possible. Also, why does the potential does NOT deppend on the radius? since we have V = Qk/r

ty vm :)

2. Aug 10, 2016

### Cozma Alex

Well, potential depends on the radius you just wrote V= kQ/ r

r...r....r....

The conditions are:

Conservation of charge and same potential on both spheres after the connection because when spheres reach the potential equilibrium there's no more flow of charges

I suggest setting a system of equation with variables the final charges on each spheres

3. Aug 10, 2016

### Fips

So what you're telling me is that when the two spheres are connected there is a flow of charge (which would be described as a function) that tends to zero as the system reaches a potential equilibrium?

4. Aug 10, 2016

### Cozma Alex

Well, you can see it like that, but i would prefer to think it using physics, like you know charges of same sign tends to repell each other, on a sphere they tend to get as far as they can till they reach equilibrium, if you connect a charged sphere to a neutral one you are actually giving the charges more space to move and make distance between each other till they reach equilibrium...

About you first question on why charges tends to be distributed not equally, you can see it also matematically:
V has to be the same at equilibrium right? If on a sphere radius is smaller to maintain the same V as the sphere with greater radius it has to change Q by smalling it, so the little sphere will also have less charges

5. Aug 10, 2016

### Fips

waaa thanks that last response was very enlightening ^^ thank you!