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

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>

> [!TLDR]
> By finding the uniform field between the plates and the potential difference, the capacitance of a parallel plate capacitor works out to C = ε₀A/d.
>
> Consider a parallel plate capacitor consisting of two large, plane parallel conducting plates, each of area $A$, separated by a small distance $d$ (with vacuum/air between them), carrying charges $+Q$ and $-Q$ respectively. Since $d$ is small compared to the linear dimensions of the plates, edge effects are neglected and the field between the plates is treated as uniform.
>
> **Step 1 — Surface charge density:**
> The surface charge density on each plate is
> $$\sigma = \dfrac{Q}{A}$$
>
> **Step 2 — Electric field between the plates:**
> Using the result (from Gauss's law) for the field due to two oppositely, uniformly charged parallel plates, the field in the region between them is uniform and given by
> $$E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A}$$
> directed from the positive to the negative plate (the fields outside the plates cancel, while the fields between the plates add up).
>
> **Step 3 — Potential difference:**
> Since the field between the plates is uniform, the potential difference between the plates, separated by distance $d$, is
> $$V = E d = \dfrac{Q d}{\varepsilon_0 A}$$
>
> **Step 4 — Capacitance:**
> Capacitance is defined as $C = Q/V$. Substituting: …

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.

Concept understanding — Capacitance of a Parallel Plate Capacitor

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