# Potential of a Charged Conducting Sphere

## Homework Statement

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.

## Homework 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

## 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)*(σ4πr^2/r^2), this gives me E*ϵ0, but this is not the correct answers. Does E=V0?

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

ehild
Homework Helper
Hi Estefania, welcome to PF!

## Homework Statement

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.

## Homework 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

## 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)*(σ4πr^2/r^2), this gives me E*ϵ0, but this is not the correct answers. Does E=V0?
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 ?

What do you mean by 0 here? Are you referring to the electric permittivity of free space, ##ε_0## ?
Yes, that's what I meant.

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

ehild
Homework Helper
The potential is given by V=Q/4pi*epsilon naught*r, so I would use this equation and set V0 equal to this?
Yes, ##V_0=\frac{Q}{4\pi\epsilon_0 r}## . (Use parentheses!)
Find Q from this.

• Estefania_8
Yes, that's what I meant.
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}##.

• Estefania_8
Yes, ##V_0=\frac{Q}{4\pi\epsilon_0 r}## . (Use parentheses!)
Find Q from this.
Yes, ##V_0=\frac{Q}{4\pi\epsilon_0 r}## . (Use parentheses!)
Find Q from this.
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!

ehild
Homework Helper
You are welcome 