Distance from potential difference

[Sho Kano]

Homework Statement

A metal sphere of radius 15 cm has a net charge of 4.2x10^-8C. At what distance from the sphere's surface has the electric potential decreased by 500V?

Homework Equations

V = kQ/r

The Attempt at a Solution

\Delta V\quad =\quad \frac { kQ }{ r } \\ \\ r\quad =\quad \frac { kQ }{ \Delta V } \\ r\quad =\quad \frac { 9e9*4.2e-8 }{ 500 } \\ =\quad 0.756\quad m

What is wrong?

 

[gneill]

You are looking for a potential difference, not a particular potential. Find the potential difference between a point located adjacent to the sphere (r_o ≅ 15 cm) and one located at some larger radius r₁. Once you know r₁ you should be able to find the required distance.

 

[Sho Kano]

You are looking for a potential difference, not a particular potential. Find the potential difference between a point located adjacent to the sphere (r_o ≅ 15 cm) and one located at some larger radius r₁. Once you know r₁ you should be able to find the required distance.

In a metal sphere the charge is distributed to the surface right?
Can I use V = kq/r for a sphere? It seems a bit weird to me, can you explain? Maybe I just don't understand the voltage equation..

 

[gneill]

In a metal sphere the charge is distributed to the surface right?
Can I use V = kq/r for a sphere? It seems a bit weird to me, can you explain? Maybe I just don't understand the voltage equation..

For a charged conducting sphere, all the charge will be located at its surface. For any spherically symmetric charge distribution the electric field external to the sphere behaves as though the total charge were a point charge located at the center of the sphere.

So yes, you can use V = kq/r for the sphere.

 

[Sho Kano]

For a charged conducting sphere, all the charge will be located at its surface. For any spherically symmetric charge distribution the electric field external to the sphere behaves as though the total charge were a point charge located at the center of the sphere.

So yes, you can use V = kq/r for the sphere.

If it's not too much trouble, why does it behave like a point charge? Is it through some kind of mathematical proof?

 

[gneill]

If it's not too much trouble, why does it behave like a point charge? Is it through some kind of mathematical proof?

Yes. It's the same method as for Newton's gravitational shell theorem.

 

[Sho Kano]

Yes. It's the same method as for Newton's gravitational shell theorem.

Sorry, for the late response. The answer that I'm getting is 0.0371 m?

 

[gneill]

Looks good. You should probably express it in cm since the sphere radius was given in those units.

 

[Sho Kano]

Looks good. You should probably express it in cm since the sphere radius was given in those units.

Good advice, the tricky part was just one of the steps involved putting d+r in the voltage equation.
Thanks for the help!