Conductor capacitance
1.
Explain the concept of "conductor capacitance"!
Solution:
2.
Derive the dimension of the physical unit farad = F. Show that it holds: C2.N-1.m-2 = F.m-1
Solution:
Thus we have shown that 1 F = A2.kg-1.m-2.s4 and that C2.N-1.m-2 = F.m-1
3.
Calculate the capacitance of a plate capacitor composed of 11 plates with dimensions 3 cm and 2 cm, if the distance between the plates is 0.2 mm. The dielectric between the plates is mica with εr = 6. There are 10 gaps between the 11 plates.
Solution:
Analysis:
S = a.b = 3cm·2cm = 6 cm2 = 6·10-4 m2, l = 2·10-4 m, εr = 6, ε0 = 8,854·10-12 C2.N-1.m2
The capacitance of the capacitor is C' = 1594 pF.
4.
What is the voltage between the plates of an air capacitor with two square plates of side 10 cm, separated by 2 cm, if its charge is 8,854·10–3 μC?
Solution:
Analysis:
S = a2 = (10 cm)2 = 100 cm2 = 10-2 m2, l = 2·10–2 m, Q = 8,854·10–9 C, ε0 = 8,854·10–12 F·m-1
5.
What is the capacitance of our Earth?
Solution:
Analysis:
R = 6378 km = 6.378·106 m, ε0 = 8,854·10-12 F·m-1, εr = 1
The Earth's capacitance is C = 709.27 μF