Q2)
Consider a capacitor consisting of two parallel conducting plates, 1 and 2, separation
a, with a dielectric material of constant relative permittivity ε between the plates.
Plate 1 is in the z = a plane and is held at a fixed potential V1, whereas plate 2 is in the
z = 2a plane and is held at the fixed potential V2. The plates are so large compared to
their separation that the electric field between them can be considered to depend only
on z. Between the plates there is a free charge density f  z   K / z 3 where K is a
constant.
i)
Show that the potential in the region a < z < 2a is,
V  z  
1 K 1  V2  V1 1 K 1 
3 K 1


z   2V1  V2  
2 
2  0 z  a
4  0 a 
4  0 a
ii)
Hence determine the electric field E in the region a < z < 2a between the plates.
(My notes)
This is a list of all the parameters that I can think of that may be useful in answering
this question, plus their symbols and equations.
Er 
Coulombs law
1
4 0

q1  r  ri 
r  ri
The complications associated with summing or integrating vectors means that it is
often more straightforward to calculate the electrostatic potential V. for a set of point
charges, the potential V is
V r 
1
4 0
3
i
q1
 rr
i
i
'
And for a charge density   r  the potential is
V r 
1
4 0
  r 
 r  r
d 
Electric displacement field relationship D =  0 E
Gauss’s law can be written in terms of D and the free charge density  f as
div D =  f
For LIH (linear, isotropic and homogeneous) materials, D =  0 E, so
div E =
1
 0
f
Which in the absence of any free charge reduces to Laplace’s equation,
 2V  0
Electrostatic potential V satisfies Poisson’s equation,
 2V  
1
 0
f
These constitute the uniqueness theorem.
I am baffled as to how to bring all these points together to answer this question, please
show me how. Thank you.