Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Potential Between Concentric Spheres, Assignments of Electrodynamics

Goes in detail to solve potential using Laplace's equation.

Typology: Assignments

2016/2017

Uploaded on 03/31/2025

jose-salazar-espitia
jose-salazar-espitia 🇺🇸

1 document

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Potential Between Concentric Spheres
We want to find the potential V(r, θ) in the region between a grounded sphere of radius aand a concentric
spherical shell of radius bheld at potential V(b, θ) = 3
2sin2θ. The region of interest is a < r < b.
1 Governing Equation and Symmetry
In the region between the conductors (a < r < b), there is no charge density (ρ= 0). Therefore, the electric
potential Vmust satisfy Laplaces equation:
2V= 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 Vwill be independent of the azimuthal angle φdue to the symmetry of
the problem (azimuthal symmetry), so V=V(r, θ). Laplaces equation in spherical coordinates with V
φ = 0
simplifies to:
1
r2
r󰀕r2V
r󰀖+1
r2sin θ
θ 󰀕sin θV
θ 󰀖= 0
Multiplying by r2gives:
r󰀕r2V
r󰀖+1
sin θ
θ 󰀕sin θV
θ 󰀖= 0
2 General Solution using Separation of Variables
The general solution to Laplaces 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 󰀓Alrl+Blr(l+1)󰀔Pl(cos θ) (1)
where Aland Blare constants determined by the boundary conditions.
3 Applying Boundary Conditions
3.1 Inner Boundary (r = a)
The inner sphere is grounded, meaning its potential is zero:
V(a, θ) = 0
Substituting r=ainto the general solution (1):
󰁛
l=0 󰀓Alal+Bla(l+1)󰀔Pl(cos θ) = 0
Since this must hold for all values of θ, and the Legendre polynomials are linearly independent (orthogonal),
the coecient of each Pl(cos θ) must be zero:
Alal+Bla(l+1) = 0
1
pf3

Partial preview of the text

Download Potential Between Concentric Spheres and more Assignments Electrodynamics in PDF only on Docsity!

Potential Between Concentric Spheres

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.

1 Governing Equation and Symmetry

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

2 ∂V

∂r

r^2 sin θ

∂θ

sin θ

∂V

∂θ

Multiplying by r 2 gives: ∂

∂r

r

2 ∂V

∂r

sin θ

∂θ

sin θ

∂V

∂θ

2 General Solution using Separation of Variables

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

  • Blr −(l+ 1 )

Pl(cos θ) ( 1 )

where Al and Bl are constants determined by the boundary conditions.

3 Applying Boundary Conditions

3. 1 Inner Boundary (r = a)

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

  • Bla −(l+ 1 )

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

  • Bla −(l+ 1 ) = 0

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 )

3. 2 Outer Boundary (r = b)

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:

  • P 0 (x) = 1
  • P 1 (x) = x
  • P 2 (x) = 1 2 ( 3 x^2 − 1 )

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):