Derive the Electrostatic Potential for Point Charges

Solve Poisson's equation for electrostatics using the method of Green's functions and visualize the equipotential surfaces for a collection of point charges

The classical formula for electrostatic potential due to a point charge is well known:

ϕ=

4π

In the formula,

is the electric potential,

is the electric charge,

is the permittivity of free space, and

is the distance from the point charge. Starting from the differential form of Gauss's Law, it is possible to derive this expression for a static point charge by using a Green's function.

Lookup the formula for reference:

In[1]:=

FormulaData["PointCharge"]

Out[1]=





4π



4π



Determining the electrostatic scalar potential is an application of Poisson's equation:

∇

ϕ=

In the above formula,

represents the electric charge density.

Define the distribution of charge density for a point charge of magnitude

located at

(

)

In[2]:=

ρ0[x_,y_,z_]:=Q*DiracDelta[x-x0,y-y0,z-z0]

Compute a Green's function for the Laplacian operator:

In[3]:=

gf=GreenFunction-

∇

{x,y,z}

ϕ[x,y,z],ϕ[x,y,z],{x,y,z}∈FullRegion[3],{a,b,c}

Out[3]=

4π

(-a+x)

(-b+y)

(-c+z)

Use the properties of Green's functions to solve the differential equation:

In[4]:=

ϕ0=Integrate[gf*ρ0[a,b,c]/

,{a,b,c}∈FullRegion[3],Assumptions{x,y,z,x0,y0,z0}∈Reals]

Out[4]=

4π

(x-x0)

(y-y0)

(z-z0)

The quantity in the denominator is simply the Euclidean distance in Cartesian coordinates:

In[5]:=

ϕ0/.(1/FullSimplify[EuclideanDistance[{x,y,z},{x0,y0,z0}],{x,y,z,x0,y0,z0}∈Reals]1/r)

Out[5]=

4πr

The standard formula has been reproduced.

Define a new distribution of charge density for two point charges, one of magnitude

located at

(

)

and another of magnitude

located at

(

)

In[6]:=

ρ[x_,y_,z_]:=q1*DiracDelta[x-x1,y-y1,z-z1]+q2*DiracDelta[x-x2,y-y2,z-z2]

Choose specific values for the charge magnitudes and positions:

In[7]:=

params={x11,y1-1,z10,x22,y20,z20,q15,q2-1,1/

4π};

Use the properties of Green's functions to solve the differential equation with this new source term:

In[8]:=

ϕExample=Integrate[(gf*ρ[a,b,c]/

)/.params,{a,b,c}∈FullRegion[3],Assumptions{x,y,z}∈Reals]

Out[8]=

(-2+x)

(-1+x)

(1+y)

Visualize the equipotential surfaces for the example charge distribution:

In[9]:=

SliceContourPlot3D[ϕExample,{x,-3,3},{y,-3,3},{z,-3,3},PlotPoints50,PlotLegendsAutomatic,ContoursRange[-5,5,.5],ColorFunction"Pastel"]

Out[9]=

Use a different plotting function to see other interior features of the equipotential surfaces:

In[10]:=

ContourPlot3D[ϕExample,{x,-3,3},{y,-3,3},{z,-3,3},RegionFunctionFunction[{x,y,z},x-2<y],RegionBoundaryStyleNone,MeshNone,ColorFunction"Pastel",Contours10,PlotLegendsAutomatic]

Out[10]=

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Derive the Electrostatic Potential for Point Charges

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Derive the Electrostatic Potential for Point Charges

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