It has only been fairly recently that 1.0 F capacitors have been readily availab

Question

It has only been fairly recently that 1.0 F capacitors have been readily available. A typical 1.0 F capaitor can withstand up to 5.00 V. To get an idea why it is not easy to make a 1.0 F capicitor, imagine making a 1.0 F parallel plate capacitor using titanium dioxide (k = 90, breakdown strength is 4.00 kV/mm) as the dielectric. Find the minimum thickness of the titanium dioxide such that the capacitor can withstand 5.00 V. Ans in m. Find the area of the plates so that the capacitance is 1.0 F. Ans in m^2.

Explanation / Answer

Capacitance of a parallel plate capacitor is C = 1.0 F Potential difference between the parallel plate capcitor is V = 5.00 V Dielectric constant for the material titanium dioxide is k = 90 Dielectric strength is E = 4.00 k V / mm                                       = 4.00 *10^3 V / 10^-3 m                                           = 4.00 *10^6 V / m thickness of the titanium oxide is d = V / E                                                         = 5.00 V /  4.00 *10^6 V / m                                                        = 1.25*10^-6 m Area of the plates of the capcitor is   C = k 0 A / d                                                    1.0F   = 90 ( 8.85*10-12 C2 / N m2 ) A / 1.25*10^-6 m                                                      A = 1.56*103 m2

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