Question

Given a fine plate with DC current injected and drained by two punctual electrodes in the sides. Is the current density greater in the edges?

Answer 1

Given a fine rectangular conducting plate (conductivity σσ), with a constant current II being injected and drained by electrodes at points (0,b)(0,b)and (2a,b)(2a,b) respectively, what is the current distribution J⃗ J→ in it?



The solution I arrived at shows a stronger current density at the edges. I expected greater density in the middle of the plate. Is the result I got correct, or maybe there is something wrong with my solution?



Method to arrive at the solution

It is considered that outside of the plate there is no conductivity, hence no current; also that all charges in the plate are free, i.e.,

∇ ˙J⃗ =0∇ ˙J→=0

considering the plate isotropic:

J⃗ =σE⃗ J→=σE→

Thus

∇ ˙E⃗ =0∇ ˙E→=0

Also, given that the current is constant, there is no time-variant magnetic field, then the electric field is irrotational and can be expressed as the gradient of a scalar field (the electric potential)

∇×E⃗ =0∇×E→=0

E⃗ =−∇uE→=−∇u

From the null divergence, we have Laplace's equation for the potential (in two dimensions):

∇2u=0∇2u=0

Boundary conditions:

u(x,0)=0u(x,0)=0

u(x,2b)=0u(x,2b)=0

∂u∂x(0,y)=Iσδ(y−b)∂u∂x(0,y)=Iσδ(y−b)

∂u∂x(2a,y)=Iσδ(y−b)∂u∂x(2a,y)=Iσδ(y−b)

δ(y)δ(y) is Dirac's delta

The solution is

u(x,y)=Iσb∑n=0∞(−1)2n+1sin(kny)cosh(kn2a)[−sinh(knx)+sinh(kn(2a−x))]knu(x,y)=Iσb∑n=0∞(−1)2n+1sin⁡(kny)cosh⁡(kn2a)[−sinh⁡(knx)+sinh⁡(kn(2a−x))]kn

Where

kn=(2n+1)π2bkn=(2n+1)π2b

Taking the gradient of uu and multiplying by (−σ)(−σ)yields the x⃗ x→ and y⃗ y→ parts of J⃗ J→

J⃗  ˙x⃗ =Ib∑n=0∞(−1)2n+1sin(kny)cosh(kn2a)[−cosh(knx)−cosh(kn(2a−x))]J→ ˙x→=Ib∑n=0∞(−1)2n+1sin⁡(kny)cosh⁡(kn2a)[−cosh⁡(knx)−cosh⁡(kn(2a−x))]

J⃗  ˙y⃗ =Ib∑n=0∞(−1)2n+1cos(kny)cosh(kn2a)[−sinh(knx)+sinh(kn(2a−x))]J→ ˙y→=Ib∑n=0∞(−1)2n+1cos⁡(kny)cosh⁡(kn2a)[−sinh⁡(knx)+sinh⁡(kn(2a−x))]

I used the Python code below to get the quiver figure above.