Question

A parallel plate capacitor is constructed using a dielectric material with dielectric constant of 3 and...

A parallel plate capacitor is constructed using a dielectric material with dielectric constant of 3 and whose dielectric strength is 2x10^8 V/m. the desired capacitance is 0.250 mF, and the capacitors must withstand a maximum potential difference of 4 kV. Find the minimum area of the capacitor plates.

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Answer #1

maximum.separation between plates of capacitor be d

Than

​d= ∆V/E ; E = dielectric strength

Hence :

d=4*103 / 2*108 =2*10-5 m

Now capacitance

\large C=\frac {k \epsilon _{o}A}{d}

\large 0.250*10^{-3}=\frac {3*8.85*10^{-12}*A}{2*10^{-5}}

A= 188m2

> when calculating d, it shouldn't be 2*10^8 instead it should be going the other way

nick kaspar Mon, Feb 7, 2022 10:53 AM

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