given a slab of material with a finite resistivity Solution a) Since the conductivities of the media above and below the sheet are negligible, no current will flow outside the slab. Furthermore, due to symmetry of the configuration, current will flow perpendicular to the injecting electrode throughout the slab. However, the strength of the current will decrease with radial distance from the point of injection. Thus, a cylindrical coordinate system can be used to conveniently describe the current distribution, with z as the perpendicular distance from the electrode and r as the radial distance. b) Let the current density be J. At a particular z, integral of J with respect to area should be equal to I (since charge can neither be created nor destroyed. We assume that J is inversely proportional to radial distance r. Thus, J = k / r At a particular z, elemental area dA = r x d(phi) (where phi is the azimuth angle of the cylindrical system) Integral of J with respect to dA = k x pi = I Therefore, k = I / pi Hence, J = I / (pi x r) From ohm\'s law, we have: E = rho x J (where rho is the resistivity) Therefore, E = rho x I / (pi x r) c) We know that E = - dV / dz Thus, Potential difference between two points will be equal to the negative integral of E with respect to z Since E depends not on z but only on r, negative integral of E with respect to z will be: V = rho x I x (d - c) / (pi x r) d) Distance z = 3 - 1 = 2 cm = 0.02 m I = 10 mA = 0.01 A Potential difference V = 0.25 V Resistance R between the two points = V / I = 25 ohms Thickness H = 5 cm = 0.05 m Resistivity rho = R x Area / Distance z = 25 x 0.05 / 0.02 = 62.5 ohm-cm .

1. given a slab of material with a finite resistivity
Solution
a)
Since the conductivities of the media above and below the sheet are negligible, no current will
flow outside the slab. Furthermore, due to symmetry of the configuration, current will flow
perpendicular to the injecting electrode throughout the slab. However, the strength of the current
will decrease with radial distance from the point of injection. Thus, a cylindrical coordinate
system can be used to conveniently describe the current distribution, with z as the perpendicular
distance from the electrode and r as the radial distance.
b)
Let the current density be J.
At a particular z, integral of J with respect to area should be equal to I (since charge can neither
be created nor destroyed. We assume that J is inversely proportional to radial distance r. Thus, J
= k / r
At a particular z, elemental area dA = r x d(phi) (where phi is the azimuth angle of the cylindrical
system)
Integral of J with respect to dA = k x pi = I
Therefore, k = I / pi
Hence, J = I / (pi x r)
From ohm's law, we have:
E = rho x J (where rho is the resistivity)
Therefore, E = rho x I / (pi x r)

2. c)
We know that
E = - dV / dz
Thus, Potential difference between two points will be equal to the negative integral of E with
respect to z
Since E depends not on z but only on r, negative integral of E with respect to z will be:
V = rho x I x (d - c) / (pi x r)
d)
Distance z = 3 - 1 = 2 cm = 0.02 m
I = 10 mA = 0.01 A
Potential difference V = 0.25 V
Resistance R between the two points = V / I = 25 ohms
Thickness H = 5 cm = 0.05 m
Resistivity rho = R x Area / Distance z = 25 x 0.05 / 0.02 = 62.5 ohm-cm