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

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Accepted Answer

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> [!TLDR]
> Using Gauss's law to find the uniform field between two oppositely charged parallel plates, then $V = Ed$ and $C = Q/V$, gives $C = \varepsilon_0 A/d$.
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> **Setup:** Consider two large, parallel conducting plates, each of area $A$, separated by a small distance $d$ (so that $d \ll$ the dimensions of the plates, allowing edge effects to be neglected). One plate carries charge $+Q$ and the other $-Q$, with surface charge density $\sigma = Q/A$.
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> **Step 1 — Electric field between the plates:**
> Each plate, being a large charged conducting sheet, produces a uniform field of magnitude $\dfrac{\sigma}{2\varepsilon_0}$ on each side (from Gauss's law applied to an infinite charged sheet). Between the plates, the fields due to the two plates point in the same direction and add up, while outside the plates they cancel. So the net field between the plates is:
> $$E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$$
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> **Step 2 — Potential difference between the plates:**
> Since the field is uniform over the separation $d$: …

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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