# How electric potential boundary condition works

Tags:
1. Jun 11, 2017

### tomasg

1. The problem statement, all variables and given/known data

Inside a sperical dielectric mass there is a electric dipole on the center of the sphere. The sphere has radius a. This dieletric sphere is inside and on the center of a conductive spherical shell of radius b. The problem asks to find the potentials and then the electric fields in every region, inside the dielectric sphere, the space between the sphere and the shell and outside the shell.

2. Relevant equations
Its given that p=p0*z (the dipole looks towards +z )

3. The attempt at a solution

Now, i have written all the potentials (the solutions of laplace) but i noticed that i havent fully understood one boundary condition for the electric potential. The one that says ε2(∂Vout/∂r)-ε1(∂V/∂r)=-σ(θ)/ε0.
The problem doesnt say anything about the charge of the shell, so i suppose is zero. So my question is this, does the σ(θ) of the above condition refers to the induced charge density (which would not be zero in this example i think) or the charge density of the shell alone?

2. Jun 12, 2017

### Govind Lal Sidhardh

well I'm not an expert so pardon me if I'm wrong...But the above equation refers to the discontinuity of the electric field in the boundaries of two media,which is derived from Gauss' law.And the charge density in Gauss' law is the net charge density(due to induction too).So I think the charge density refers to the net charge density,which in this case is the induce charge density...

3. Jun 12, 2017

### tomasg

Thanks for the reply. I believe this is true too. Atleast thats what i have understand after reading more caerefully griffith's book

4. Jun 12, 2017

### TSny

This equation does not look quite right. The left side gives the change in the radial component of D (not E). The right hand side should then be -σfree(θ) without any ε0.

σfree is the free charge density on the boundary surface between the two dielectric materials.

If you wrote the left hand side without the ε2 and ε1 as ∂Vout/∂r- ∂V/∂r, then you now have radial components of E. Then the right hand side would be -σ(θ)/ε0 and σ(θ) would be the total surface charge density (free plus bound).

I have an older 3rd edition of Griffiths. In this edition, there is a section 4.3.3 called "Boundary Conditions". Here you find the equations

Dabove - Dbelow = σf where σf is free charge density.

and

Eabove - Ebelow = σ/ε0 where σ is total charge density.

Last edited: Jun 12, 2017
5. Jun 13, 2017

### tomasg

yes i should not put the ε0 there. Thank you so much sir for the answer. It finally makes sense to me. And as the user above said, the Eabove - Ebelow = σ/ε0= (σf+σb)/ε0 <--- in this equation the σf could be the induced charge density or the charge density we created in the conductor (or both). Right?

6. Jun 14, 2017

### TSny

I'm not sure I'm understanding your question. In your problem, the inner surface of the conducting sphere will have a free charge density σf which is the charge density induced on the inner surface of the spherical conductor by the dielectric sphere with the dipole. So if you are applying the equation to the inner surface of the conductor, the σ in σ/ε0 would be this σf.

Last edited: Jun 14, 2017