Statement
Examine the coaxial cones shown in figure (a) below. Assume a potential V1 on the inner cone, and a zero potential reference on the outer cone. The cone vertices are insulated at r = 0. The space between the conductors is free space. (a) Solve Laplace's equation for the region between the cones. (b) Find the voltage at an angle between the cones, and conversely the angle at a voltage between the cones. (c) Find the charge distribution on the outer cone, and the capacitance between the cones.
System Parameters
Permittivity of free space:
Solution
The potential on the cones is constant with r and f, by inspection, since there is no discontinuity in charge density on either cone. Laplace's equation reduces to the q component only.
(a)
Integrating:
P.26 |
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P.27 Laplace Equation in Spherical Coordinates |
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Select the variable q in the expression below and choose Integrate on Variable from the Symbolic menu. (If you are using the Mathcad Engine, you won't have this option.)
yields
plus a constant.
Select the expression for V and choose Simplify to get
or
(Remember that ln(a) - ln(b) = ln(a/b).)
Solve for the variables A and B using a solve block.
Guess:
For more information on solve blocks, see A Mathcad Tutorial.
(b)
To find the voltage at some angle between the cones, create the formula
In reverse, to find an angle for a given voltage, create the expression
where
select the q, and choose Solve for Variable from the Symbolic menu. (If you are using the Mathcad Engine, you won't have this option.)
(c)
To find the charge density on the outer cone, solve for the flux density via the electric field. Also, let's change the angle of the outer cone to 90º, so that it becomes a flat plane.
Guess:
Thus,
since
The electric field is found by taking the negative gradient of V:
Choose Simplify from the Symbolic menu (if you are using the Mathcad Engine, you won't have this option) to yield
where the field points in the q direction, that is, it arcs between the two cones. The flux density is
where
The direction of D (into the plane) means that the normal vector is opposite in sign, and that the surface charge is negative.
where
A graph of the charge density distribution
To find the capacitance between the cones, neglect fringing, and choose new angles:
Guess:
The electric field, as was shown earlier, is
so
And the charge density on the upper plate is
Since we know the distance to the edge of the cones this time (see second figure above), it's possible to calculate the total charge on the upper plate. The radius is given by the angle between the cones and the distance defined in the picture.
The capacitance is
Remember, when adjusting the starting parameters for this problem, you may need to try new values for the solve block guesses.
