Two conductors exist in the x-y plane. One of the conductors is a
right angle formed from the +x and +y axis, and extends for a distance
of 50cm along each axis as shown. The second conductor exists along
the hyperbola xy = (4*100) cm2 in the first quadrant between the lines
x = 50 cm and y = 50cm. An electric field E = 100[ ^x(-y) +
^y(-x) ] V/m exists in the free space region between the two
conductors.

Note: ^x means x hat. ^y means y hat.

Note: The electric field has a unit in term of volt per meter, not centimeter

a) Show that the electric field satisfies the equation of
electrostatics in free space based on Maxwell?s equation.

b) Using the boundary conditions on the right-angle conductor, find
the total charge on the conductor surface on the yz plane and on the
xz plane for length 100 cm in the z direction.

c) Perform a line integral of the electric field between the two
conductors to find the potential difference between them.

d) Find the capacitance of this device for length 100 cm in the z
direction using the results obtained in questions b and c.

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a)	Check that the Curl of E = 0.  
   For E = 100 (-y,-x) in vector notation.  Take the cross products of
the vectors:  (d/dx, d/dy, d/dz)  and (-y, -x, 0), where d/dx etc are
partial derivatives.
      -dx/dx +dy/dy = -1 + 1 = 0.

     Thus the equation of electrostatics in free space is satisfied.


b)	Assume that the conductors extend to infinity in the z direction. 
We will compute the charge density first and then find the total
charge.

Find the electric potential, phi:

phi = - Integral[ E*ds ] =  - 100 Integral[-y*dx] - 100 Integral[-x*dy]
    = 200xy + Vo

Where Vo is a constant.


We employ the method of conformal mapping, as described on p.737  of
Riley, Hobson and Bence, ?Mathematical Methods for Physics and
Engineering,? 2nd Edition Cambridge University Press, 2003.  Since
conformal mapping preserves the orthoganality of field lines and
potentials, it allows us to transform or unfurl the problem into a
simple parallel-plate capacitor problem. In this new transformed space
we can easily solve for the charge and the voltage.


We want an analytic function of the complex plane w = x + i*y whose
real part is phi, namely: f(w)= phi + i*psi

This is our complex potential.


To find the complex potential we need find psi, so we use the
Cauchy-Riemann relations:
dpsi/dy = dphi/dx = 200y   
dpsi/dx = -dphi/dy = -200x

Integrating, 
psi = 100y^2 =100x^2 + constant

Letting Vo absorb our integration constant, we plug psi and phi into f(w):

f(w)= Vo + 200xy + 100i[y^2-x^2]

Written in terms of w, (recall w=x+iy) our complex potential simplifies to:
    f(w) = -i*100w^2   

Apply the conformal map, m=w^(1/2)  This unfolds our L-shaped
conductor into a straight line along the  x-axis, straightening the
curved conductor as well.

Let F(m) be the complex potential in our conformed space:

F(m)= f(w) = -i*100m + Vo

Let PHI be our electric potential in our conformed space:

PHI = Re[ F(w) ] = Re[ -i*100(x+iy)+ Vo ] = 100y + Vo

Now we find the electric field in our conformed space, E?:
E? =  - grad(PHI) = -100 * y_hat   

In our conformed space, we have a constant electric field directly
above our now-straight conductor. Pretty nifty!

We compute the charge density, sigma, by using the equation which
governs an infinite conducting sheet of charge.  This is a good
approximation, since the length of the conductors is large compared to
the distance between them.

E? = sigma / epsilon0 

sigma = E? *epsilon0 = -100 * epsilon0 coulombs / meter^2

Since charge is conserved, even under conformal mapping, the total
charge Q on the L-shaped plates extended 1 meter in z is:
 Q= 1 meter * 1 meter * sigma  = -100 * epsilon0  coulombs of charge.

 Epsilon0= 8.854 *10^-12 

So the total charge: Q ~= -8.854 * 10 ^-10 Coulombs 


c)	We will take the line integral in the conformed space.  First we
note that in the original space, the conductor which lies along the
curve xy = .04 rests upon an equipotential line:

phi = 8 + Vo  Volts

In the conformed space, our conductor will still lie on the
transformed equipotent line whose value is 8 + Vo Volts.

phi = Phi   <=>  8 + Vo = 100y + Vo
In the new space the conductor lies on y =.08
The formerly curvy conductor is now a straight line too!


The potential difference is the line integral between the two conductors:

PHI = -Integral[E*dy] for y=0 to y=.08 = 8 Volts 



d)	Using the equation for a capacitor:
C = Q/(delta V)

Capacitance is C = -100 * epsilon0 / 8 Farads
C ~= 1.1 *10 ^-10 Farads