Derive an expression for the capacitance of a parallel plate capacitor.

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> [!TLDR]
> Using Gauss's law to find the field between the plates and then integrating to get the potential difference, the capacitance of a parallel plate capacitor works out to $C = \varepsilon_0 A/d$.
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> Consider a parallel plate capacitor made of two large, plane, parallel conducting plates, each of area A, separated by a small distance d, with vacuum (or air) between them. Let the plates carry charges $+Q$ and $-Q$, so the surface charge density is $\sigma = Q/A$.
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> **Step 1 — Field between the plates:** Using Gauss's law, the field due to an infinite charged plane sheet of surface density $\sigma$ is $E = \sigma/2\varepsilon_0$ on each side. Between the two oppositely charged plates, the fields due to both plates add up (they point in the same direction in the region between the plates), while outside the plates they cancel. So the net field between the plates is:
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> $E = \dfrac{\sigma}{2\varepsilon_0} + \dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A}$
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> **Step 2 — Potential difference:** Since the field between the plates is uniform, the potential difference between the plates is:
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Q.Derive an expression for the capacitance of a parallel plate capacitor.

Concept understanding — Capacitance of a Parallel Plate Capacitor

Q.Derive an expression for the capacitance of a parallel plate capacitor.

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