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Goes in detail to solve potential using Laplace's equation.
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We want to find the potential V (r, θ) in the region between a grounded sphere of radius a and a concentric
spherical shell of radius b held at potential V (b, θ) = 3 2 sin
2 θ. The region of interest is a < r < b.
In the region between the conductors (a < r < b), there is no charge density (ρ = 0 ). Therefore, the electric
potential V must satisfy Laplace’s equation:
∇ 2 V = 0
Due to the spherical geometry and the θ-dependent boundary condition on the outer shell, we use spherical
coordinates (r, θ, φ). The potential V will be independent of the azimuthal angle φ due to the symmetry of
the problem (azimuthal symmetry), so V = V (r, θ). Laplace’s equation in spherical coordinates with ∂V ∂φ
simplifies to:
1
r^2
∂r
r
∂r
r^2 sin θ
∂θ
sin θ
∂θ
Multiplying by r 2 gives: ∂
∂r
r
∂r
sin θ
∂θ
sin θ
∂θ
The general solution to Laplace’s equation in spherical coordinates with azimuthal symmetry is found using
the method of separation of variables. It is given by an infinite series involving Legendre polynomials,
Pl(cos θ):
V (r, θ) =
l= 0
Alr l
Pl(cos θ) ( 1 )
where Al and Bl are constants determined by the boundary conditions.
The inner sphere is grounded, meaning its potential is zero:
V (a, θ) = 0
Substituting r = a into the general solution ( 1 ):
l= 0
Ala l
Pl(cos θ) = 0
Since this must hold for all values of θ, and the Legendre polynomials are linearly independent (orthogonal),
the coefficient of each Pl(cos θ) must be zero:
Ala l
Solving for Bl:
Bl = −Al
al
a−(l+^1 )^
= −Ala 2 l+ 1
Substitute this relationship back into the general solution:
V (r, θ) =
l= 0
Alr l − Ala 2 l+ 1 r −(l+ 1 )
Pl(cos θ)
V (r, θ) =
l= 0
Al
r l − a 2 l+ 1 r −(l+ 1 )
Pl(cos θ) ( 2 )
The potential on the outer shell is given as:
V (b, θ) =
sin 2 θ
We need to express this boundary condition in terms of Legendre polynomials. Recall the first few Legendre
polynomials with x = cos θ:
We want to express sin 2 θ = 1 − cos^2 θ = 1 − x^2. From the definition of P 2 (x):
P 2 (cos θ) =
( 3 cos 2 θ − 1 ) =⇒ 2 P 2 (cos θ) = 3 cos 2 θ − 1
3 cos 2 θ = 2 P 2 (cos θ) + 1 = 2 P 2 (cos θ) + P 0 (cos θ)
cos 2 θ =
P 2 (cos θ) +
P 0 (cos θ)
Now substitute this into the expression for sin 2 θ:
sin 2 θ = 1 − cos 2 θ
= P 0 (cos θ) −
P 2 (cos θ) +
P 0 (cos θ)
P 0 (cos θ) −
P 2 (cos θ)
So the boundary condition becomes:
V (b, θ) =
P 0 (cos θ) −
P 2 (cos θ)
V (b, θ) = P 0 (cos θ) − P 2 (cos θ)
4 Determining the Coefficients (Al)
Equate the general solution evaluated at r = b (Equation ( 2 )) with the boundary condition expressed in
Legendre polynomials:
l= 0
Al
b l − a 2 l+ 1 b −(l+ 1 )
Pl(cos θ) = P 0 (cos θ) − P 2 (cos θ)
By comparing the coefficients of Pl(cos θ) on both sides (using orthogonality):