Parallel Plate Air Capacitor→
A parallel-plate capacitor consists of two long, plane, metallic plates mounted on two insulating stands and placed at a small distance apart in a vacuum (or air). The plates are exactly parallel to each other.
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| Parallel Plate Air Capacitor |
Derivation of the capacitance of parallel plate capacitor in the air→
Let us consider, Two plates $X$ and $Y$ are separated at a small distance $d$ in the vacuum (or air). If the area of plates is $A$ and the plates $X$ and $Y$ have charge $+q$ and $-q$ respectively. If the surface charge density on each plate is $\sigma$ then electric field intensity at a point between two parallel plates is
$E=\frac{\sigma}{\epsilon_{\circ}}$
$E=\frac{q}{\epsilon_{\circ}A}\qquad(1) \qquad (\because \sigma=\frac{q}{A}) $
The potential difference between the parallel plates capacitor in air (or vacuum) is
$V=E.d$
Now substitute the value of $E$ from equation$(1)$ in the above equation
$V=\frac{q}{\epsilon_{\circ}A}.d \qquad(2)$
The capacitance of the parallel plate capacitor in air (or vacuum) is
$C_{\circ}=\frac{q}{V} \qquad(3)$
From equation $(2)$ and equation $(3)$
$C_{\circ}=\frac{q}{\frac{q}{\epsilon_{\circ}A}.d}$
$C_{\circ}=\frac{\epsilon_{\circ}A}{d}$
If the space between the plates is filled with some dielectric medium of dielectric constant $K$, then the electric field between the plates is
$E=\frac{\sigma}{K \: \epsilon_{\circ}}$
The Potential difference between the parallel plates when dielectric medium $(K)$ is present between them
$V=\frac{q}{K \: \epsilon_{\circ} \: A}.d $
Now, The capacitance of the parallel-plate capacitor when the dielectric medium $(K)$ is present between them
$C=\frac{K\: \epsilon_{\circ} \: A}{d}$
