The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.70 and a dielectric strength of 1.90×10^7 V/m . The capacitor is to have a capacitance of 1.05×10^−9 F and must be able to withstand a maximum potential difference of 5600 V.(adsbygoogle = window.adsbygoogle || []).push({});

What is the minimum area the plates of the capacitor may have? use 8.85*10^-12 for permittivity of free space.

A=______ m^2.

Well i know that the capacitance equals permittivity times area times dialectric constant divided by distance between plates. The distance i got by dividing the voltage by the dialectric strength. Is that correct?

the final answer i got was 9.47*10^11 but it was wrong. Help please!

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# Homework Help: Dielectric with a parallel-plate capacitor finding minimum area

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