# Potential in a sphere in a dielectric

• nagyn
In summary: It's confusing since E0 has another important meaning.So, what is E in the region outside the conducting sphere?In summary, we are trying to determine the potential at the center of a conducting sphere embedded in a neutral nonconducting shell with a dielectric constant of 3.0. Using Gauss's law, we find that the dielectric has no effect on the electric potential outside of itself. Therefore, the potential at infinity remains zero. Integrating the electric field from infinity to R, we find that the potential at the center of the conducting sphere is kq/R, and with the dielectric in place, it becomes kq/2KR. To maintain a potential of zero at infinity, the potential in the sphere
nagyn

## 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 2 R . The material of which the shell is made has a dielectric constant of 3.0.

Relative to a potential of zero at infinity, what is the potential at the center of the conducting sphere?

## Homework Equations

V = ∫Edr = kq/r + C

Gauss's law in dielectrics: ∫KEdA = q/ε0

## The Attempt at a Solution

The field in a conducting sphere is zero, which means the potential in a conducting sphere is constant. Looking at just the sphere alone, potential at the center relative to infinity should be kq/R (positive because I'm going against electric field?).

With the dielectric in place to still take zero at infinity I need to take into account the potential difference across the dielectric (I think?)

Edielectric = q/KAε0

V(R->2R) = -∫Edielectric*dr (negative because I'm going in direction of electric field?) = -kq/Kr [2R - R] = -kq/2KR + kq/KR = kq/2KR

I know that a dielectric will decrease the electric potential, so I thought V = Vsphere - Vdielectric = kq*((1/R) - (1/6R)).

This is not correct and I'm not sure where I messed up.

I agree with your work up to this point:
nagyn said:
so I thought V = Vsphere - Vdielectric = kq*((1/R) - (1/6R)).
Think through that again. I believe you have made quite a simple mistake in what to subtract from what.

Maybe more straightforward if you just integrated the E field from infinity to R.

rude man said:
Maybe more straightforward if you just integrated the E field from infinity to R.

So integral of E field from infinity to R = (integral of E field from infinity to 2R) + (integral of E field through dielectric).

From from infinity to 2R, can I still consider the system as a point particle and say the electric field is kQ/R^2? Or does that change with the dielectric in place?

nagyn said:
So integral of E field from infinity to R = (integral of E field from infinity to 2R) + (integral of E field through dielectric).

From from infinity to 2R, can I still consider the system as a point particle and say the electric field is kQ/R^2? Or does that change with the dielectric in place?
Answer this by applying Gauss' law to a Gaussian surface of radius r>2R.

EA = Qenclosed/E0

If I'm understanding correctly, a dielectric doesn't change the total charge enclosed, so Qenclosed should still be Q.

E = Q/AE0 = kQ/R^2

nagyn said:
EA = Qenclosed/E0

If I'm understanding correctly, a dielectric doesn't change the total charge enclosed, so Qenclosed should still be Q.

E = Q/AE0 = kQ/R^2
Yes, but I'm not sure why this thread has become so complicated.
A dielectric diminishes the potential difference across it. It exerts no long range field, so cannot affect potential gradients outside itself, as you confirmed by following TSny's hint.
In post #1 you found the dielectric layer reduced the potential difference across it from kq/2R to kq/2KR. What reduction in potential in the sphere is needed to keep the potential at infinity zero?

nagyn said:
EA = Qenclosed/E0
If I'm understanding correctly, a dielectric doesn't change the total charge enclosed, so Qenclosed should still be Q.
E = Q/AE0 = kQ/R^2
good thought!
but don't say "E0" when you mean ε0.

## What is potential in a sphere in a dielectric?

Potential in a sphere in a dielectric refers to the amount of electrical potential energy per unit charge that exists within a charged sphere surrounded by an insulating material, also known as a dielectric.

## How is potential in a sphere in a dielectric calculated?

The potential in a sphere in a dielectric can be calculated using the equation V = Q/(4πεr), where V is the potential, Q is the charge on the sphere, ε is the permittivity of the dielectric material, and r is the radius of the sphere.

## What factors affect potential in a sphere in a dielectric?

The potential in a sphere in a dielectric is affected by the charge on the sphere, the permittivity of the dielectric material, and the distance between the center of the sphere and the surrounding material.

## How does potential change in different types of dielectric materials?

The potential in a sphere in a dielectric can change depending on the type of dielectric material used. Materials with a higher permittivity will result in a higher potential, while materials with a lower permittivity will result in a lower potential.

## Why is potential in a sphere in a dielectric important in scientific research?

Potential in a sphere in a dielectric is important in scientific research because it helps to understand the behavior of charges and electric fields in insulating materials. This knowledge is crucial in a wide range of fields, including electronics, material science, and engineering.

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