# How is voltage on a sphere affected by edges?

• G Cooke

#### G Cooke

I have two metal half-spheres, each with a hole in the center. If I put these two together, perhaps with electrical tape, and apply a charge, how will the voltage due to the nonuniform charge distribution be different from that of a perfect sphere?

I am assuming that the charge distribution will be nonuniform due to the edges present at the holes and possibly at the seam between the two halves since charge is always more dense at edges and corners, so V will not be exactly kq/r.

I am mainly interested in whether the voltage would be greater or less than kq/r and how significant the difference would be. Although I wouldn't mind it if someone actually came up with a new function for the voltage.

You'll probably need a two-dimensional numerical simulation for a proper estimate. I would not expect large deviations as long as the holes are small compared to the sphere diameter, probably (that is a guess) increasing the voltage (relative to infinity) a bit.

It could be interesting to study the limiting case of holes that cover nearly the whole half-spheres, so your setup becomes ring-like.

G Cooke
The voltage will be the same throughout the metal as that is the nature of a conductor in equilibrium state. You are correct in that the charge density will not be uniform. There will be slightly more dense charge near the boundary. Since you are "squeezing" more charge in a smaller area (having cut out a hole) you will have a slightly higher value than kq/r for the voltage.

G Cooke
You'll probably need a two-dimensional numerical simulation for a proper estimate. I would not expect large deviations as long as the holes are small compared to the sphere diameter, probably (that is a guess) increasing the voltage (relative to infinity) a bit.

It could be interesting to study the limiting case of holes that cover nearly the whole half-spheres, so your setup becomes ring-like.
Thank you for your reply! The holes are small compared to the diameter.

So the seam between the two half-spheres wouldn't have any effect?

The voltage will be the same throughout the metal as that is the nature of a conductor in equilibrium state. You are correct in that the charge density will not be uniform. There will be slightly more dense charge near the boundary. Since you are "squeezing" more charge in a smaller area (having cut out a hole) you will have a slightly higher value than kq/r for the voltage.
Thank you for your reply! That does make sense that since we are packing more charge in a smaller surface area than if the hole were filled, we will have a larger voltage.

So you don't think the seam between the spheres would have any effect?

So the seam between the two half-spheres wouldn't have any effect?
Depends on details of the seam.
That does make sense that since we are packing more charge in a smaller surface area than if the hole were filled, we will have a larger voltage.
Well, you also have a smaller area... but still the same radius.

G Cooke
Very little effect if the seams are not deviating far from the spherical shape. What is typically the difficulty with sharp edges is that at a given high voltage the charge density gets high enough that breakdown occurs and you get currents off the surface (arching or ionization). But it is not the voltage per se which again is the same throughout the metal, but rather the off-surface potential gradient and the charge density that causes issues with sharp edges and such.

G Cooke
Depends on details of the seam.
I assume that by details you are referring to any deviations from perfect flushness all the way around. But I suppose, as jambaugh said, these will have little effect if they are small.
Well, you also have a smaller area... but still the same radius.
I think you're thinking about the case where you put charge on a sphere and then cut out a hole. In that case, you couldn't simply say that because the area is smaller, the voltage is larger. But if there is already a hole and you put on some charge, it must necessarily have a greater voltage than if there were no hole, right?

Last edited:
I assume that by details you are referring to any deviations from perfect flushness all the way around. But I suppose, as jambaugh said, these will have little effect if they are small.
Right.
I think you're thinking about the case where you put charge on a sphere and then cut out a hole.
Well then you reduce the total charge and the situation is quite obvious.
But if there is already a hole and you put on some charge, it must necessarily have a greater voltage than if there were no hole, right?
I would expect that, but I don't see a fully convincing argument yet.

G Cooke
Well, you also have a smaller area... but still the same radius.
Oh, I see what you're saying now. You're saying that even though the presence of a hole means that we're putting the same charge in a smaller area, the radius is still the same, so kq/r will still be the same whether there's a hole or not.
I would expect that, but I don't see a fully convincing argument yet.
I would argue that voltage is charge over capacitance, and capacitance is by definition a function of only conductor surface area, space between conductors (undefined for a single sphere), and permittivity of the dielectric inside the conductor (assumed to be air); therefore, if given the amount of charge, voltage is solely dependent on (and inversely proportional to) surface area. For a sphere, surface area itself depends solely on radius; therefore voltage can be said to depend solely on radius (hence V = kq/r). But the introduction of a hole parameter terminates the surface area's sole dependence on radius; therefore, voltage cannot be said to depend solely on radius, and V = kq/r no longer applies. The new formula would likely involve the radius of the hole along with the sphere radius.

In short, V = kq/r applies only for actual spheres.

Last edited:
Capacitance in general is not proportional to surface area. It applies to some special geometries, but it is not true in general.
In short, V = kq/r applies only for actual spheres.
Right.

G Cooke