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 the plates, then V = Ed and C = Q/V, gives C = ε₀A/d.
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> **Setup**
> Consider a parallel plate capacitor made of two large, plane, parallel conducting plates, each of area $A$, separated by a small distance $d$ (small compared to the plate dimensions, so edge effects are negligible), with vacuum (or air) between them. Let the plates carry charges $+Q$ and $-Q$, with uniform surface charge density $\sigma = Q/A$.
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> **Step 1: Field due to one charged plate**
> Using Gauss's law, the electric field due to a single infinite plane sheet of charge with surface density $\sigma$ is:
> $$E_{	ext{sheet}} = \frac{\sigma}{2\varepsilon_0}$$
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> **Step 2: Field between the plates**
> Between the two oppositely charged plates, the fields from the positive plate and the negative plate both point from the positive plate toward the negative plate, so they **add up**:
> $$E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A\varepsilon_0}$$
> (Outside the plates, the two fields cancel, so $E \approx 0$ there.)
> …

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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