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For electrolysis and computation of resistances of grounding plates, we have a conductive medium with conductivity σ and a steady current. The current density J is related to the electric field E through J = σE. Combining the continuity equation ∇ · J = Q, where Q is a current source, with the definition of the electric potential V yields the elliptic Poisson's equation
–∇ · (σ∇V) = Q.
The only two PDE parameters are the conductivity σ and the current source Q.
The Dirichlet boundary condition assigns values of the electric potential V to the boundaries, usually metallic conductors. The Neumann boundary condition requires the value of the normal component of the current density (n · (σ∇V)) to be known. It is also possible to specify a generalized Neumann condition defined by n · (σ∇V) + qV = g, where q can be interpreted as a film conductance for thin plates.
The electric potential V, the electric field E, and the current density J are all available for plotting. Interesting quantities to visualize are the current lines (the vector field of J) and the equipotential lines of V. The equipotential lines are orthogonal to the current lines when σ is isotropic.
Two circular metallic conductors are placed on a plane, thin conductor like a blotting paper wetted by brine. The equipotentials can be traced by a voltmeter with a simple probe, and the current lines can be traced by strongly colored ions, such as permanganate ions.
The physical model for this problem consists of the Laplace equation
–∇ · (σ∇V) = 0
for the electric potential V and the boundary conditions:
V = 1 on the left circular conductor
V = –1 on the right circular conductor
The natural Neumann boundary condition on the outer boundaries
$$\frac{\partial V}{\partial n}=0.$$
The conductivity σ = 1 (constant).
Open the PDE app by typing
pdetool
at the MATLAB^{®} command prompt.
Click Options > Application > Conductive Media DC.
Click Options > Grid Spacing..., deselect the Auto check boxes for X-axis linear spacing and Y-axis linear spacing, and choose a spacing of 0.3, as pictured. Ensure the Y-axis goes from –0.9 to 0.9. Click Apply, and then Done.
Click Options > Snap
Click and draw the blotting paper as a rectangle with corners in (-1.2,-0.6), (1.2,-0.6), (1.2,0.6), and (-1.2,0.6).
Click and add two circles with radius 0.3 that represent the circular conductors. Place them symmetrically with centers in (-0.6,0) and (0.6,0).
To express the 2-D domain of the problem, enter
R1-(C1+C2)
for the Set formula parameter.
To decompose the geometry and enter the boundary mode, click .
Hold down Shift and click to select the outer boundaries. Double-click the last boundary to open the Boundary Condition dialog box.
Select Neumann and click OK.
Hold down Shift and click to select the left circular conductor boundaries. Double-click the last boundary to open the Boundary Condition dialog box.
Set the parameters as follows and then click OK:
Condition type = Dirichlet
h = 1
r = 1
Hold down Shift and click to select the right circular conductor boundaries. Double-click the last boundary to open the Boundary Condition dialog box.
Set the parameters as follows and then click OK:
Condition type = Dirichlet
h = 1
r = -1
Open the PDE Specification dialog box by clicking PDE > PDE Specification.
Set the current source, q, parameter to 0.
Initialize the mesh by clicking Mesh > Initialize Mesh.
Refine the mesh by clicking Mesh > Refine Mesh twice.
Improve the triangle quality by clicking Mesh > Jiggle Mesh.
Solve the PDE by clicking .
The resulting potential is zero along the y-axis, which is a vertical line of anti-symmetry for this problem.
Visualize the current density $$J$$ by clicking Plot > Parameters, selecting Contour and Arrows check box, and clicking Plot.
The current flows, as expected, from the conductor with a positive potential to the conductor with a negative potential.
The Current Density Between Two Metallic Conductors