2. a) Assume you have four circular plates: two of them are ofradius of 10mm and
ID: 1872665 • Letter: 2
Question
2. a) Assume you have four circular plates: two of them are ofradius of 10mm and the other two are of radius of 20mm. Choose two plates from above to design an air-filled parallel-plate capacitor with a capacitance of 1pF with the minimal distance between the plates. Explain your choice and calculate the plate separation. i) If there is a potential of 10V applied across the plates, what is the charge value on the plates, and how much energy can be stored in this capacitor? ii) Discuss how the stored energy can be enhanced at the same applied voltage. v) Choose one dielectric material from Table 2-a as the filling dielectric material between the plates in your design, to achieve a capacitance close to 300pF Dielectric Material Teflon Polyimide Titanium Dioxide Strontium Titanate Barium Strontium Titanate 500 2.1 3.4 86 310Explanation / Answer
1) We will choose two plates of radius 10 mm or 0.01 m. With this and the plate separation of around 2.78 mm or 2.78e-3, we can achieve the required capacitance of 1 pF or 1e-12 F with minimal plate separation. The formula used was C = e0 * A / d where A is area of plate and d is plate sepation
C = 8.85e-12*3.142e-4 / 2.78e-3
C = 1E-12 OR 1pF (required capacitance)
2) Q = C*V = 1E-12*10 = 1E-11 coulombs of charge
Now, energy stored = 1/2*C*V2 = 0.5*1e-12*100 = 5e-11 Joules
3) By increasing the capacitance, we can enhance the stored energy while applying the same voltage , and to increase the capaitance we can either increase the plate area or decrease the plate separation.
4) The capacitanc formula with dielectric medium is C = Er*e0*A/d
Let x be the required Er . Then Er = C*d / e0*A
It will come as 300 and from the given table it is closest to Strontium Titanate.
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