A parallel plate, air filled capacitor, has plates of area 0.82 m2 and a separat

Question

A parallel plate, air filled capacitor, has plates of area 0.82 m2 and a separation of 0.060 mm. (a) Find the capacitance. nF (b) If a voltage of 25 volts is applied to the capacitor, what is the magnitude of the charge on each plate? C (c) What is the energy stored in the capacitor? J Now consider an identical capacitor, however, the space between this capacitor contains strontium titanate which has a dielectric constant of 310 and a dielectric strength of 8.0x106 N/C. (d) What is the new value of capacitance? F (e) For a voltage of 25 V, what is the charge on each plate? C (f) What is the energy stored in the capacitor now? J (g) What is the maximum voltage that can be placed across the capacitor? V

Explanation / Answer

a) C = (e0*Area)/*(distance between plates) = 121.00nF b)q= C*V=3025.18nC c) Energy stored = U = (1/2)*C*V^(2) =0.5*q*V= 3.7812*10^(-5)J d)C = (e0*K*Area)/*(distance between plates) = 37.5µF e)q = CV = 937.5µC f)U = (1/2)*C*V^(2) =0.11718J g)Vmax=E*d=8.0*10^6 *0.06*10^(-3)=480 V **I am not sure about your data for dielectric field strength ...so i have assumed it 8.0*10^6 rather than 8*106...but if your data is right than just replace 8.0*10^6 to 8*106.. with

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