# Potential of a Charged Conducting Sphere

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1. Apr 4, 2015

### Estefania_8

1. The problem statement, all variables and given/known data
A conducting sphere with radius R is charged to voltage V0 (relative to a point an infinite distance from the sphere where the potential is zero). What is the surface charge density σ? Express your answer in terms of the given quantities and ϵ0.

2. Relevant equations
Electric field on a sphere, E=(1/4πϵ0)*(Q/r^2)
Surface charge density: σ=Q/A, where Q is the charge and A is the area of the surface.
A for a sphere: A=4πr^2

3. The attempt at a solution
I need to find σ and I have Q in two of the equations I replaced A with the area of a sphere, so I get
σ=Q/4πr^2
Then I solved for Q from this, so I have Q=σ4πr^2, now I substitute this into the first equation and solve for σ, so now I have, E=(1/4πϵ0r^2)*(σ4[I]πr^2/r^2), this gives me E*[I]ϵ[/I]0, [/I]but this is not the correct answers. Does E=V0?

2. Apr 4, 2015

### PWiz

What do you mean by 0 here? Are you referring to the electric permittivity of free space, $ε_0$ ?

3. Apr 4, 2015

### ehild

Hi Estefania, welcome to PF!

You need to give σ in terms of V0 , the potential of the sphere. What is the formula for the potential of a charged sphere ?

4. Apr 4, 2015

### Estefania_8

Yes, that's what I meant.

5. Apr 4, 2015

### Estefania_8

The potential is given by V=Q/4pi*epsilon naught*r, so I would use this equation and set V0 equal to this?

6. Apr 4, 2015

### ehild

Yes, $V_0=\frac{Q}{4\pi\epsilon_0 r}$ . (Use parentheses!)
Find Q from this.

7. Apr 4, 2015

### PWiz

Then you have all the variables you'll need. Follow ehild's procedure and eliminate $Q$ from the equations. This should be easy as you've already stated yourself that $σ=\frac Q{4πr^2}$.

8. Apr 4, 2015

### Estefania_8

OK, so now I have V0*4πε0*r=Q, so now I would just use σ=Q/A, which would translate into σ=Q/4πr^2, like PWiz said. Then I would use σ4πr^2=Q, which would give me V0*4πε0*r=σ4πr^2, so solving for σ I get Voε0/r=σ and that is the answer. Thank you to you both!

9. Apr 4, 2015

### ehild

You are welcome