# Electric Potential at Center of Sphere

• lightofthemoon
In summary, the potential at the center of a charged conducting sphere is the same as the potential just outside of the sphere, but the dielectric affects the potential at infinity, so the potential at the sphere (kq/R) - new potential at infinity should be .5 (because of dielectric constant).
lightofthemoon

## Homework Statement

A solid conducting sphere of radius R and carrying charge +q is embedded in an electrically neutral nonconducting spherical shell of inner radius R and outer radius
9 R . The material of which the shell is made has a dielectric constant of 2.0.
Relative to a potential of zero at infinity, what is the potential at the center of the conducting sphere?

V=kq/r
k=1/4πϵ0
κ=V0/Vd

## The Attempt at a Solution

V=kq/R
Potential at center of conducting sphere should be the same as the potential just outside of the sphere
Since the dielectric constant = 2
V0/κ = Vd
I think the answer should be V=kq/2R ?

However I don't think this is the correct answer... I'm not really sure how to use the outer radius in my calculation...

Remember that potential is always relative to a chosen zero. In the present case, you are told to take the potential at infinity as zero. So you need to find how the dielectric affects the potential at infinity.

Since the potential at infinity is told to be zero, shouldn't it still be zero no matter how the dielectric affects it? If I use the equation
V0/κ = Vd , 0 / 2 = 0...

lightofthemoon said:
Since the potential at infinity is told to be zero, shouldn't it still be zero no matter how the dielectric affects it? If I use the equation
V0/κ = Vd , 0 / 2 = 0...
Start without the dielectric. You just have a charged conducting sphere. There is a standard formula for the potential at the sphere, and this corresponds to a zero potential at infinity.
Now we add the dielectric shell. If we take the potential at the charged sphere to be the same as before, we find the potential at infinity has changed. So in order to get the potential at infinity back to zero we must add or subtract something from the potential at the sphere.

I don't really think I understand what you are saying.

So standard potential at the sphere is kq/R
and potential at infinity is 0

So when you add the dielectric shell the potential at infinity will increase because there is less potential because of the dielectric shell? So it should be potential at sphere (kq/R) - new potential at infinity which will be .5 (because of dielectric constant) kq/9R, which equals kq/R - kq/18R or 17kq/18R ?

Last edited:
lightofthemoon said:
I don't really think I understand what you are saying.

So standard potential at the sphere is kq/R
and potential at infinity is 0

So when you add the dielectric shell the potential at infinity will increase because there is less potential because of the dielectric shell? So it should be potential at sphere (kq/R) - new potential at infinity which will be .5 (because of dielectric constant) kq/9R, which equals kq/R - kq/18R or 17kq/18R ?
Nearly right. I don't think the kq/9R is right, since that is based on kq/R at the sphere.
You know the field through the dielectric shell. Compute the drop in potential from inside to outside.

haruspex said:
Nearly right. I don't think the kq/9R is right, since that is based on kq/R at the sphere.
You know the field through the dielectric shell. Compute the drop in potential from inside to outside.
Why isn't it based off of the kq/R at the sphere? Not sure it should be if I don't use the sphere as a basis... it is kq/8R?

lightofthemoon said:
Why isn't it based off of the kq/R at the sphere
As I wrote, potential is always relative to a chosen zero. The usual formula of kq/R for a sphere is obtained by taking the potential at infinity (with no intervening dielectric) as zero. A more generic formula would be kq/R+c, where c is whatever constant is necessary to fit with the chosen zero potential. You can think of it as a constant of integration (integrating the field).

Essentially the same problem was poted by nagyn on 02feb at 0557 GMT.

## 1. What is the electric potential at the center of a sphere?

The electric potential at the center of a sphere can be calculated using the formula V = kQ/R, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the sphere, and R is the radius of the sphere. The result will be in units of volts (V).

## 2. How does the electric potential at the center of a sphere change with distance?

The electric potential at the center of a sphere follows an inverse relationship with distance. This means that as the distance from the center of the sphere increases, the electric potential decreases.

## 3. Can the electric potential at the center of a sphere be negative?

Yes, the electric potential at the center of a sphere can be negative if the charge of the sphere is negative. This indicates that the sphere is surrounded by a negatively charged field.

## 4. What factors affect the electric potential at the center of a sphere?

The electric potential at the center of a sphere is affected by the charge of the sphere, the distance from the center of the sphere, and the medium surrounding the sphere. It also depends on the presence of other charged objects in the vicinity.

## 5. How does the electric potential at the center of a sphere differ from the electric field at the center of a sphere?

The electric potential at the center of a sphere is a measure of the potential energy of a unit charge at that point, while the electric field at the center of a sphere is a measure of the force exerted on a unit charge at that point. They are related, but not the same, as the electric field is the derivative of the electric potential.

• Introductory Physics Homework Help
Replies
2
Views
250
• Introductory Physics Homework Help
Replies
7
Views
5K
• Introductory Physics Homework Help
Replies
22
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
292
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
23
Views
860
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
21
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
2K