 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 $1.60×{10}^{7}\frac{V}{m}$. The capacitor is to have a capacitance of $1.25×{10}^{-9}F$ 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

Expert

Given
We are given the dielectric constant $K=3.60$. The dielectric strength is the electric field that the dielectric could tolerate before breakdown is occur and equals $E=1.60×{10}^{7}\frac{V}{m}$. Also, we are given the capacitance of the capacitor $C=1.25×{10}^{-9}F$ and the potential difference $V=5500V$
Required
We are asked to find the minimum area A of the plates.
Solution
The capacitor withstands till the electric field equals $1.60×{10}^{7}\frac{V}{m}$, after this a breakdown occurs. This electric field is related to the distance d between the two plates
$E=\frac{V}{d}$
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 ${E}_{max}$ and V, we can calculate the minimum distance ${d}_{min}$​​​​​​​ as next
${d}_{min}=\frac{V}{{E}_{max}}$
$=\frac{5500V}{1.6×{10}^{7}\frac{V}{m}}$
$=3.44×{10}^{4}m$
The minmum area is related to the minimum distance as given by equation 24.19 in the next form
$A=\frac{Cd}{K{ϵ}_{0}}$
Where ${ϵ}_{0}$, is the electric constant and equals $8.854×{10}^{-12}\frac{{C}^{2}}{N}\cdot {m}^{2}$ (See Appendix F) and C is the capacitance of the capcitor in the presence of the dielectric. Now we can plug our values for C, K, d and ${ϵ}_{0}$, into equation (1) to get the minimum area A
$A=\frac{Cd}{K{ϵ}_{0}}$
$=\frac{\left(1.25×{10}^{-9}F\right)\left(3.44×{10}^{-4}m\right)}{\left(3.60\right)\left(8.854×{10}^{-12}\frac{{C}^{2}}{N}\cdot {m}^{2}\right)}$
$=0.0135{m}^{2}$
$=135c{m}^{2}$ Neunassauk8

Expert

The capacitance with dielectric with dielectric constant K is $C=\frac{AK{ϵ}_{0}}{d}$ and maximum
potential difference , ${V}_{max}={E}_{max}d$
Thus, area of plate $A=\frac{Cd}{K{ϵ}_{0}}=\frac{C{V}_{max}}{K{ϵ}_{0}{E}_{max}}=\frac{\left(1.25×{10}^{-9}\right)\left(5500\right)}{3.60×\left(8.85×{10}^{-12}\right)\left(1.60×{10}^{7}\right)}=0.0135{m}^{2}$

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