Question
Derive an expression for capacitance of the parallel plate capacitor.
✓
Answer
$\rightarrow$ As shown in figure, the charge on plate 1 and 2 are $+Q$ and $-Q$ respectively. And ' d ' is the separation between them, A is the area of each plate.
$\rightarrow$'d' is much smaller than the linear dimension of the plates $[d \ll A]$ so we can use the result on electric field $E =\sigma / 2 \varepsilon_0$ by an infinite plane sheet of uniform surface charge density
where, $\sigma=$ surface charge density, $\pm \sigma= Q / A$
$\varepsilon_0=$ permittivity of space
$A=$ Area of plate, $Q=$ charge on plate
$\rightarrow A s=$ shown in figure in region (1),
$E=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0 \ldots(1)$
$\rightarrow$ In region (2)
$E=\frac{\sigma}{2 \varepsilon_0}-\frac{\sigma}{2 \varepsilon_0}=0 \ldots(2)$
$\rightarrow$ Electric field between plate 1 and 2 ,
$E=\frac{\sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0}=\frac{\sigma}{\varepsilon_0}=\frac{\sigma}{\varepsilon_0 A}\ldots(3)$
$\rightarrow$ For uniform electric field,
$\begin{aligned}
V & =E \cdot d \\
& =\frac{Q}{\varepsilon_0 A} \cdot d \ldots(4)
\end{aligned}$
$\rightarrow$ Capacitance $C=\frac{Q}{V}=\frac{Q}{\frac{Q d}{\varepsilon_0 A}}=\frac{\varepsilon_0 A}{d}\ldots(5)$
$C=\frac{\varepsilon_0 A}{d}$ is the equation of parallel plate capacitor.
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