The problem involves calculating the distance from the surface of a charged metal sphere at which the electric potential decreases by a specified amount. The subject area is electrostatics, focusing on electric potential and charge distribution in conductive materials.
Some participants have provided guidance on how to approach the problem by clarifying the need to find the potential difference between two points. There is an ongoing exploration of the underlying principles of electric fields and potential in relation to spherical charge distributions.
Participants note the importance of unit consistency, as the radius of the sphere is given in centimeters, which may affect the final expression of the distance calculated.
In a metal sphere the charge is distributed to the surface right?gneill said:You are looking for a potential difference, not a particular potential. Find the potential difference between a point located adjacent to the sphere (ro ≅ 15 cm) and one located at some larger radius r1. Once you know r1 you should be able to find the required distance.
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.Sho Kano said: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..
If it's not too much trouble, why does it behave like a point charge? Is it through some kind of mathematical proof?gneill said: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.
Yes. It's the same method as for Newton's gravitational shell theorem.Sho Kano said:If it's not too much trouble, why does it behave like a point charge? Is it through some kind of mathematical proof?
Sorry, for the late response. The answer that I'm getting is 0.0371 m?gneill said:Yes. It's the same method as for Newton's gravitational shell theorem.
Good advice, the tricky part was just one of the steps involved putting d+r in the voltage equation.gneill said:Looks good. You should probably express it in cm since the sphere radius was given in those units.