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In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{-3} \mathrm{~m}^2$ and the distance between the plates is $3 \mathrm{~mm}$. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 $V$ supply, what is the charge on each plate of the capacitor?
Solution:
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Verified Answer
Given, $A=6 \times 10^{-5} \mathrm{~m}^2, d=3 \mathrm{~mm}$
$$
\begin{aligned}
&=3 \times 10^{-3} \mathrm{~m}, \varepsilon_0=8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\
&\mathrm{~V}=100 \mathrm{~V}, \mathrm{C}=?, \mathrm{q}=?
\end{aligned}
$$
By formula, $\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$
$$
\begin{aligned}
&=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} \\
&=17.718 \times 10^{-12} \mathrm{~F}=17.7 \mathrm{pF} .
\end{aligned}
$$
And by relation, $\mathrm{q}=\mathrm{CV}=17.7 \times 10^{-12} \times 100$ $=17.7 \times 10^{-10} \mathrm{C}$.
$$
\begin{aligned}
&=3 \times 10^{-3} \mathrm{~m}, \varepsilon_0=8.854 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\
&\mathrm{~V}=100 \mathrm{~V}, \mathrm{C}=?, \mathrm{q}=?
\end{aligned}
$$
By formula, $\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$
$$
\begin{aligned}
&=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} \\
&=17.718 \times 10^{-12} \mathrm{~F}=17.7 \mathrm{pF} .
\end{aligned}
$$
And by relation, $\mathrm{q}=\mathrm{CV}=17.7 \times 10^{-12} \times 100$ $=17.7 \times 10^{-10} \mathrm{C}$.
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