
1. How many electrons per
cm^{2} do you need on the surface a capacitor made of parallel
metal plates in air with an area of 1 cm^{2} and distance 1
cm to provide for some field E ending there? 

The relation between the field E
resulting from a homogeneous twodimensional charge distribution and the charge
density r_{area} is 


E = 
Q_{ }
e_{0} · A 
= 
r_{area}
e_{0} 




with Q = charge in [C], A
= area considered, r_{area} =
areal density of the charge. 


Compare the two formulas for the capacity C of
the capacitor formed by the parallel plates if you are not sure about the
equation above. We have: 


C = 
e_{0} ·A
d_{ } 
= 
Q
U 
and 
E = 
U
d 






2. What would
be the maximum density for reasonable field strengths up to an ultimate limit
of about 10 MV/cm? (For higher field strengths you will get violent
discharge). 

Lets look at some numbers (e_{0} = 8,854 · 10^{–14}
C/Vcm) 


Field strength 
10^{3} V/cm
(rather low) 
10^{5} V/cm
(breakdown limit
of "normal dielectrics" 
10^{7} V/cm
(close to ultimate breakdown limit) 
r_{area} 
8,85 · 10^{–11}
C/cm^{2} 
8,85 · 10^{–9}
C/cm^{2} 
8,85 · 10^{–7}
C/cm^{2} 


3. How does
this number compare to the average volume density of electron in metals.
Consider, e.g., from how far away from the surface you have to collect
electrons to achieve the required surface density, if you allow the volume
density in the afflicted volume to decrease by x %? 

The average volume density of
electrons in metals is about 1 electron/atomic volume. 


Lets keep thing easy and take for the size
d_{atom} = 1 Å, which gives 1
Å^{3} = 10^{–3} nm^{3} for the volume of
one atom. The volume density of atoms or
electrons per cm^{3} is thus r_{volume} = 10^{24}
electrons/cm^{3}. 


The areal density is
whatever is contained in a volume with an extension of just one atom diameter
in one direction, i.e. 


r_{areal} = 
r_{volume} ·
d_{atom} 
= 10^{–17} electrons/cm^{2} 




If we want to collect a surplus charge of
Q_{surplus} = 8,85 · 10^{–7}
C/cm^{2}, the maximum charge from above, from a volume
V_{surplus} by reducing the concentration of
10^{–24} electrons/cm^{3} by x %, we
have 


Q_{surplus} = 
r_{volume} · d
100 · x 
d = 
100 · x · Q_{surplus}
r_{volume} 
= 
x · 8,85 · 10^{–5}
10^{24} 
cm 
= 
x · 8,85 · 10^{–29} 
cm 



In words: Whatever value we like for
x, we only have to change the volume concentration of the electrons in
an extremely thin layer a tiny little bit to produce any areal charge densities needed  in metals, that is! 

