A large capacitance of 140 mF is needed for a certain applic

A large capacitance of 1.40 mF is needed for a certain application. Calculate the area the parallel plates of such a capacitor must have if they are separated by 3.50 mum of Teflon. What is the maximum voltage that can be applied? Find the maximum charge that can be stored. Calculate the volume of Teflon alone in the capacitor. Find the total capacitance of the combination of capacitors shown in Figure 18.24. (c_1=22.0 muF, c_2=2.50 muF.) bank with a total capacitance of 0.654 F and you possess numerous 1.50 mF capacitors. What is the smallest number you could book together to achieve your goal?

Solution

For a capacitor we know that the capacitance is given as: C = _orA/d where _o is the permittivity of air and _r is the relative permittivity of the dielectric. Also, A is the area of the plates and d is the separation between them

Part a.) Using the relation above we get:

C = 1.4 x 10^-3 = 8.85 × 10^-12 x 2.1 x A / 3.5 x 10^-6

or, A = 1.4 x 10^-3 x 3.5 x 10^-6 / 8.85 × 10^-12 x 2.1 = 0.264 10^3 = 264 m²

Part b.) The dielectric strength of teflon is 60 MV/m

So for a capacitor with distance between the plates as 3.5 um, the maximum voltage the capacitor can have would be: V = 60 x 10^6 x 3.5 x 10^-6 = 210 Volts

Part c.) The maximum charge would be CVmax = 1.4 x 10^-3 x 210 = 0.294 C

Part d.) Volume of teflon would be: Ad = 264 x 3.5 x 10^-6 = 924 x 10^-6 m³

Second Question:

We know that for capacitors C1 and C2, connected in parallel the net capacitor is given as: Cnet = C1 + C2 while for capacitors connected in series, the net capacitor is C1 x C2 / C1 + C2

So for the given circuit we can write: C₁₂ = C1 + C2 = 24.5 F

Hence the net capacitance = Cnet = 0.3 x 24.5 / 24.8 = 0.296 F