A certain substance has a dielectric constant of 2.8 and a dielectric strength o

Question

A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18.0 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 6.00×10-8 F and to ensure that the capacitor will be able to withstand a potential difference of 4.4 kV?

Explanation / Answer

1)to find area of plates C=k*e0*A/d 2)you need to find plate separation so V=Ed => d=V/E so d=4.4*10^3V/18*10^6 V/m = distance between plates 3) use the distance between plates to find A: C=k*e0*A/d => A=Cd/ke0 = [(6*10^-8)*(distance between plates above)]/[2.8*(8.85*10^-12)] = area in meters squared i dont have a calculator so you will have to find the distance between the plates and plug it into the second equation, you will see where to put it.

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