Electrostatic Boundary Condition & Laplace Equation | Electricity & Magnetism - Physics PDF Download

Electrostatics Boundary Conditions 

The boundary between two mediums is a thin sheet of surface charge σ . Consider a thin Gaussian pillbox, extending equally above and below the sheet as shown in the figure below:
The Gauss's law states that,

The normal component of  is discontinuous by an amount σ/ε_0, at any boundary. If there is no surface charge, E^⊥.
The tangential component of  is always continuous.

Apply  to the thin rectangular loop,where  stands for the components of  parallel to the surface.The boundary conditions on  can be combined into the single formula:where  is the unit vector perpendicular to the surface, pointing upward.The potential is continuous across any boundary sincethe path shrinks to zero.

where  denotes the normal derivative of V (that is the rate of change in the direction perpendicular to the surface).

Example 1: Assume that the z = 0 plane is the interface between two linear and homogenous dielectrics (see figure). The relative permittivities are ε_r = 5 for z > 0 and ε_r = 4 for z < 0. The electric field in the region z > 0 is  

If there are no free charges on the interface, then find an electric field in the region z < 0.

Solution of Laplace’s equation for simple cases

Since,  and 

This is known as Poisson's equation.

In regions where there is no charge, so that ρ=0 , Poisson's equation reduces to Laplace's equation,

Example 2: Consider two concentric spherical conducting shells centred at the origin. The outer radius of the inner shell is r_a and the inner radius of the outer shell is r_b . The charge density ρ = 0 in the region r_a < r < r_b. If V = 0 at r = ra and V = V₀ at r = r_b, then find V in the region r_a < r < r_b.

Since voltage is varying only with r, the Laplace’s equation takes the form
Integrate twice to get the solution V (r ) = A ln ( r ) + Band the boundary conditions are  (i) V = 0 at r = r_a 
(ii) V = V₀ at r = r_b 
Substituting these boundary conditions, we get
Thus,

Example 3: Potential in a region of space is given by,  where φ₀ and a is constant. Then find the charge density in this region.

Example 4: If the electrostatic potential was given by φ =φ₀ ( x²+ y² +z²), where φ₀ is constant, then find the charge density giving rise to the above potential.

∇²φ = -ρ/ε₀,

where ∇² is the Laplacian operator and ε₀ is the vacuum permittivity.

Given φ = φ₀ (x² + y² + z²), we have:

∇²φ = ∂²φ/∂x² + ∂²φ/∂y² + ∂²φ/∂z² = 2φ₀ + 2φ₀ + 2φ₀ = 6φ0.

Now, equating this to -ρ/ε₀, we get:

-ρ/ε₀ = 6φ₀

Therefore, the charge density ρ is given by:

ρ = -6ε₀φ₀

So, the charge density giving rise to the given potential is -6ε₀φ₀