Problem Set 5 Group

Problem G5-1

Editor's Note:  The following solution was presented by the Buttermilk group.  It has been edited for format and content.

Let's visualize the capactor:

Clear["Global`*"]

<< Geometry`Rotations`

\(SquareMaker[θ_] := 
Show[
        Graphics[
          Line[{Rotate2D[{\(-1\), \(-1\)}, N[θ]], 
              Rotate2D[{1, \(-1\)}, N[θ]], 
              Rotate2D[{1, 1}, N[θ]], 
              Rotate2D[{\(-1\), 1}, N[θ]], 
              Rotate2D[{\(-1\), \(-1\)}, 
                N[θ]]
}]], 
Graphics[
          Line[{
{\(-1\), \(-1\)}, {\(-1\), 1}, {1, 
                1}, {1, \(-1\)}, {\(-1\), \(-1\)}
}]], 
Graphics[Text["A", { .6, 1.2}], 
          TextStyle -> {FontFamily -> "Helvetica", \ 
              FontSize -> 24}], 
        Graphics[Text["C", {0,  .85}], 
          TextStyle -> {FontFamily -> "Helvetica", \ 
              FontSize -> 14}], 
        Graphics[Text["B", {\(- .3\), 1.2}], 
          TextStyle -> {FontFamily -> "Helvetica", \ 
              FontSize -> 14}], 
        Graphics[Text["θ", {\(- .1\), 1.05}], 
          TextStyle -> {FontFamily -> "Helvetica", \ 
              FontSize -> 14}], 
AspectRatio -> 
          Automatic]\)

SquareMaker[π/3]

The side of the square (a) is equal to the sum of the sides of the triangle; a=A+B+C.

It is apparent that all the triangles are equal.  These define the sides of the triangle:

B = c Cos[θ];
A = c Sin[θ];

Plugging these into C=a-A-B, we get the area as the Square minus the 4 Triangles:

c = a/(1 + Sin[θ] + Cos[θ]);

Area = (A + B + c)^2 - 4 (A B/2)

area[θ_] = Simplify[Area]

Clear[a]

a = 3;
Plot[Area, {θ, 0, Pi/2},  
      PlotRange -> {0, a^2}]

Since rotating the capacitor by π/2 takes us back to the original configuration, we only need to show rotation from 0 to π/2, and assume periodicity after that.

We are neglecting edge effects , and assuming the only important area of the capacitor is where the squares overlap;
Now let's define a function for internal energy.  For a parallel plate capacitor U=QV/2 where Q=σ A and V=Q C where C=(Subscript[ϵ, 0])A/L
C is capacitance, V is potential, Q is charge, and σ is charge density.  Normalized:

σ = 1; (Subscript[ϵ, 0]) = 1; a = 1;

U = 
      Area L σ^2/(2 (Subscript[ϵ, 0])) // Simplify

Plot3D[U, {θ, 0, Pi/2}, {L, 0, 2}, 
      AxesLabel -> {θ, L, "Stored Energy"}, 
      ViewPoint -> { .75, -1, 0}]

\(The\ Attractive\ Force\ goes\ as\ 1\/r\^2, \ 
    which\ in\ this\ case\ is\ vertical\ distance\ as\ we\ pull\ the\ plates\ apart, since\ we\ are\ not\ moving\ them\ parallel\ to\ each\ other, \ 
    plus\ the\ change\ in\ Energy\ of\ the\ Capacitor\ with\ separation\ \(\((dU\/dL)\)\(.\)\)\)

Force = Simplify[D[U, L]]  +  1/(L^2)

Plot3D[Force, {θ, 0, Pi/2}, {L,  .1, 5}, 
      AxesLabel -> {θ, L, "Force"}, 
      ViewPoint -> {1, 1,  .5}]

ContourPlot[Force, {θ, 0, Pi/2}, {L,  .1, 5}, 
      ColorFunction -> Hue]

The 3D graph shows us that much greater Force is required when the plates are near to each other, and much less farther out.
The Contour Plot shows that rotation is more important at greater L, this is because the attractive force has fallen off considerably, and the highest change on Force left is to adjust the overlapping area- the θ term.

The Tourque is the change in energy as the capacitor is rotated.  We can ignore the attractive force, because it acts parallel to L (τ=F×R):

Torque = D[U, θ]

Plot3D[Torque, {θ, 0, Pi/2}, {L,  .1, 5}, 
      AxesLabel -> {θ, L, "Torque"}]

The Torque increases as the capacitor is turned (θ).  The torque also increases with L.