Variable induced surface charge density on a sphere

[Apashanka]

Homework Statement

Homework Equations

While solving this problem at r >>a ,the corresponding potential due to the dipole is kpcosθ/r²(potential due a dipole) where k is the electrostatic constt. ...(1)

If σ(θ) is the surface charge density induced due to external electric field.
then the dipoel moment of the configuration is 2π∫σ(θ)a³sinθ cosθ dθ from 0 to π...(2)
Now comparing (1) with the right most term of potential given in question the dipole moment is p=4πε₀a³
And comparing this with (2)
4ε₀=∫σ(θ)sin2θdθ from 0 to π.
Now the charge density at 30° is to be calculated .
will anyone please help me in sort out this.as I got stuck here.

The Attempt at a Solution

 

[TSny]

My hint would be to recall the relationship between σ and the electric field E for conductors in electrostatic equilibrium.

 

[Apashanka]

My hint would be to recall the relationship between σ and the electric field E for conductors in electrostatic equilibrium.

Yes that E=σ(θ)/ε₀ near the conductor and kpcosθ/r² far away from conductor where p is the dipole moment.
But then how to proceed??
Very near to the conductor surface is σ(θ)/ε₀ radially and E₀ along k

 

[TSny]

Yes that E=σ(θ)/ε₀

Good. Find E at the surface of the sphere from the expression for $\phi(r, θ)$ that is given.

 

[Apashanka]

Good. Find E at the surface of the sphere from the expression for $\phi(r, θ)$ that is given.

But this potential is valid only for large distances ,that's why the induced charges on the sphere is considered to be dipole and hence the potential is due to a dipole,but for distances near the sphere the potential will not be the same.

 

[TSny]

But this potential is valid only for large distances

I believe it's correct for any point outside or on the sphere. I checked a couple of standard textbooks.

 

[Apashanka]

I believe it's correct for any point outside or on the sphere. I checked a couple of standard textbooks.

But it's given in the question that for r>> and also the potential form is coming to be same which is due to external field and dipole polarisation of the sphere (∝1/r²),
Near the sphere the there will not be any 1/r² potential dependence.

 

[TSny]

But it's given in the question that for r>>a

I wasn't sure how to interpret the notation

. I'm not familiar with the symbol between the $r$ and the $a$.

But, I still claim that the expression as given holds everywhere outside the sphere, including very near the surface and even on the surface of the sphere.

In the standard derivation, you use the boundary condition for very large $r$ in order to eliminate some of the terms in the infinite series expansion of $\phi(r, \theta)$. But, the fact that you use a boundary condition for $r \gg a$ doesn't imply that the solution only holds for $r \gg a$. Maybe this is not what you were thinking, but I am wondering why you feel that the given expression for $\phi(r, \theta)$ only holds for $r \gg a$.

and also the potential form is coming to be same which is due to external field and dipole polarisation of the sphere(∝1/r²)
Near the sphere the there will not be any 1/r² potential dependence.

What is the reason for stating this? The field outside the sphere due to the induced surface charge will be exactly like that of a dipole located at the center of the sphere.

 

[Apashanka]

Good. Find E at the surface of the sphere from the expression for $\phi(r, θ)$ that is given.

Yes at the surface the electric field is 3kpcosθ/a³=σ(θ)/ε₀
where k is the electrostatic constt.
and dipole moment p=4πε₀a³E₀
I got it now.
Thank you very much

 

[TSny]

OK. Good work.