Laplace equation

We solve the Laplace-Poisson problem in cylindrical coordinates to find the variation of electrostatic force and potential with radial distance between two concentric cylinders. We need to solve Laplace's equation with the boundary condition that the potential is held constant (V(l)=0 at the inner cylinder, i.e, the filament of a diode, and V(d)=V0 at the anode cylinder). No charges are present between the cylinders, i.e., Poisson eq. reduces to a Laplace eq. The situation is different, in principle, if the electron plasma in a diode is taken into account, i.e., once the filament is heated and a current flows. The electronic charge cloud develops its own potential which can impede the flow of current (space-charge-limited region of the current-voltage characteristic). We ignore this effect here, i.e., we consider only the potential due to the external voltage.

We need to solve the equation given in Cartesian coordinates as

diff(V,x$2)+diff(V,y$2)+diff(V,z$2) = 0

We make the approximation that the cylinder is very long, and therefore the potential does not depend on the height z .

The problem reduces to one dimension in cylindrical coordinates.

> restart; with(linalg):

Warning, the protected names norm and trace have been redefined and unprotected

Maple can remind us of the Laplacian in various orthogonal coordinate systems (without explicit computations). We are interested in the radial part only:

> v1 := [r, theta, z]:
laplacian(f(r), v1, coords=cylindrical);

(diff(f(r),r)+r*diff(f(r),`$`(r,2)))/r

> DE:=diff(V(r),r$2)+diff(V(r),r)/r=0;

DE := diff(V(r),`$`(r,2))+diff(V(r),r)/r = 0

This differential equation has a very simple solution:

> dsolve(DE,V(r));

V(r) = _C1+_C2*ln(r)

The boundary conditions of zero potential at the filament and potential V0 at the cylinder can be imposed, but the thickness of the filament has to be taken into account due to the log-divergence at r =0.

> BC:=V(l)=0,V(d)=V0;

BC := V(l) = 0, V(d) = V0

> sol:=dsolve({DE,BC},V(r));

sol := V(r) = V0*ln(l)/(ln(l)-ln(d))-V0*ln(r)/(ln(l...

> Vp1:=simplify(rhs(sol));

Vp1 := -V0*(-ln(l)+ln(r))/(ln(l)-ln(d))

We enter the equivalent expression by hand:

> Vp:=V0*ln(r/l)/ln(d/l);

Vp := V0*ln(r/l)/ln(d/l)

This expression is more natural, as it makes it obvious that the logs are taken of dimensionless numbers. One way to read the equation is to say that we are measuring the distances r and d as multiples of the filament radius l .

Maple will not recognize immediately that the two expressions are identical:

> simplify(Vp-Vp1);

V0*(ln(r/l)*ln(l)-ln(r/l)*ln(d)-ln(d/l)*ln(l)+ln(d/...

> #assume(l>0,d>0,r>0); # with this command activated the line below simplifies to 0.

> simplify(Vp-Vp1);

V0*(ln(r/l)*ln(l)-ln(r/l)*ln(d)-ln(d/l)*ln(l)+ln(d/...

The electric field decreases inversely proportional with the radial distance (as expected from the divergence of the field lines). Its radial component is calculated from:

> E:=-diff(Vp,r);

E := -V0/(r*ln(d/l))

We graph the electric field using as a distance scale the filament radius l , and assume that the anode cylinder has a radius d of 100 times the filament radius.

> plot(subs(l=1,d=100,V0=100,E),r=1..100);

[Maple Plot]

We verify that this is just a graph of -1/ r (up to some factor) by graphing a log-log plot (the negative of E has to be graphed so that the log can be taken):

> with(plots):

Warning, the name changecoords has been redefined

> loglogplot(subs(l=1,d=100,V0=100,-E),r=1..100,scaling=constrained);

[Maple Plot]

The graph below shows how the voltage increases as a function of radial distance.

> plot(subs(l=1,d=100,V0=100,Vp),r=1..100);

[Maple Plot]

This result generalizes a simple result known from the parallel-plate condenser: when a voltage V 0 is applied across two parallel plates, the electric field between the plates is constant (the field lines are parallel, i.e., the field is homogeneous) and the electric potential changes linearly as one goes from one plate to the other. In the case of the cylindrical geometry the situation is different: the field lines follow radial beams that connect the two circular cross sections, i.e., the distance between them increases as one goes from the inner cylinder to the outer one. Correspondingly, the electric field weakens with radial distance. We note that the electric field falls like 1/ r (as opposed to being constant), and that the potential grows only logarithmically (rather than linearly) in this different geometry.

The following graph illustrates the cylinders (filament and anode cylinder) via their cross sections.

> with(plottools):

> c1 := disk([0,0], 1, color=red):
c2 := circle([0,0], 10, color=blue):

> for i from 1 to 20 do: phi:=2*Pi*i/20: l1[i]:=line([1*cos(phi),1*sin(phi)],[10*cos(phi),10*sin(phi)],color=green): od:

> display(c1,c2,seq(l1[i],i=1..20),scaling=constrained,axes=boxed);

[Maple Plot]

If we need to calculate the electric field in Cartesian coordinates, we substitute and differentiate:

> VpC:=subs(r=sqrt(x^2+y^2),Vp);

VpC := V0*ln(sqrt(x^2+y^2)/l)/ln(d/l)

> [-diff(VpC,x),-diff(VpC,y)];

[-V0*x/((x^2+y^2)*ln(d/l)), -V0*y/((x^2+y^2)*ln(d/l...

One can see that the force is central (in the x - y plane): the orientation of the electric field is towards the center, i.e., proportional to the position vector r = [ x , y ]. The force on a charged particle of charge q is given as F = q E .

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