maduregimc

2022-01-10

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of $1.60\times {10}^{7}\frac{V}{m}$ . The capacitor is to have a capacitance of $1.25\times {10}^{-9}F$ and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

reinosodairyshm

Beginner2022-01-11Added 36 answers

Given

We are given the dielectric constant

Required

We are asked to find the minimum area A of the plates.

Solution

The capacitor withstands till the electric field equals

As shown, the electric field is inversely proportional to the distance d and E represents the maximum electric field between the two plates before the breakdown, therefore the distance d represents the minimum distance between the two electrodes before a breakdown. As we are given

The minmum area is related to the minimum distance as given by equation 24.19 in the next form

Where

Neunassauk8

Beginner2022-01-12Added 30 answers

The capacitance with dielectric with dielectric constant K is $C=\frac{AK{\u03f5}_{0}}{d}$ and maximum

potential difference ,${V}_{max}={E}_{max}d$

Thus, area of plate$A=\frac{Cd}{K{\u03f5}_{0}}=\frac{C{V}_{max}}{K{\u03f5}_{0}{E}_{max}}=\frac{(1.25\times {10}^{-9})\left(5500\right)}{3.60\times (8.85\times {10}^{-12})(1.60\times {10}^{7})}=0.0135{m}^{2}$

potential difference ,

Thus, area of plate

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