Parallel Plate Capacitors are made up of electrodes as well as an isolating substance or substrate, and its area calculation can be defined as follows:
Given:
[tex]\bold{Q= 2.8 \ \mu}\\\\\bold{V= 2.8 \ V}\\\\\bold{d=1.0\ mm}\\\\[/tex]
To find:
Area=?
Solution:
Using formula:
[tex]\to \bold{Q = CV}\\\\\to \bold{C=\frac{Q}{V}}\\\\[/tex]
[tex]\bold{= \frac{2.8\times 10^{-6}}{200}} \\\\\bold{= 1.4\times 10^{(-8)}\ \ F}[/tex]
[tex]\bold{\therefore}\\\\\to \bold{e_0=8.85 \times 10^{-12}}[/tex]
[tex]\to \bold{C = e_0\times \frac{A}{d}}\\\\\to \bold{A=\frac{C \times d}{e_0}}[/tex]
[tex]\bold{= \frac{1.4\times 10^{-8} \times 1.0 \times 10^{-3} } {8.85\times 10^{-12}}} \\\\\bold{= \frac{1.4\times 1.0 \times 10^{-11} } {8.85\times 10^{-12}}} \\\\\bold{= \frac{1.4 \times 10^{-11} } {8.85\times 10^{-12}}} \\\\\bold{= \frac{1.4 \times 10^{-11} } {8.85\times 10^{-12}}} \\\\\bold{=0.158 \times 10} \\\\\bold{=1.58\ m^2} \\\\[/tex]
Therefore the area "1.58 [tex]\bold{m^2}[/tex]".
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