# Solution to Exercise 2.5.1

| 1. How many electrons per cm2 do you need on the surface a capacitor made of parallel metal plates in air with an area of 1 cm2 and distance 1 cm to provide for some field E ending there?
| The relation between the field E resulting from a homogeneous two-dimensional charge distribution and the charge density rarea is
E = | Q e0 · A | = | rarea e0
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|  | with Q = charge in [C], A = area considered, rarea = 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 = | e0 ·A d | = | Q U | and | E = | U d
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| 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 (e0 = 8,854 · 10–14 C/Vcm)
Field strength | 103 V/cm (rather low) | 105 V/cm (breakdown limit of "normal dielectrics" | 107 V/cm (close to ultimate breakdown limit)
rarea | 8,85 · 10–11 C/cm2 | 8,85 · 10–9 C/cm2 | 8,85 · 10–7 C/cm2
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| 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 datom = 1 Å, which gives 1 Å3 = 10–3 nm3 for the volume of one atom. The volume density of atoms or electrons per cm3 is thus rvolume = 1024 electrons/cm3.
|  | The areal density is whatever is contained in a volume with an extension of just one atom diameter in one direction, i.e.
rareal = | rvolume · datom | = 10–17 electrons/cm2
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|  | If we want to collect a surplus charge of Qsurplus = 8,85 · 10–7 C/cm2, the maximum charge from above, from a volume Vsurplus by reducing the concentration of 10–24 electrons/cm3 by x %, we have
Qsurplus = | rvolume · d 100 · x
--- | ---
d = | 100 · x · Qsurplus rvolume | = | x · 8,85 · 10–5 1024 | cm | = | x · 8,85 · 10–29 | cm
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| 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!

With frame

Exercise 2.5.1

Exercise 2.4.1

© H. Föll (Advanced Materials B, part 1 - script)