A parallel plate capacitor has rectangular plates each of area $20 c m^{2}$ is separated by a distance D the potential difference between the two plates?

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Solution

Area $= 2 c m^{2} = \frac{20}{10000} m^{2}$
$= \frac{2}{1000} m^{2}$
Let distance $= D$ , charge $= Q$
$C = \frac{(ϵ_{o}) A}{D} = \frac{ϵ_{o} 2 / 100}{D}$
$↓$
Capacitance
$P D = V = Q / C$
$= Q / \frac{ϵ_{o} \times 2 / 100}{D}$
$\Rightarrow \frac{1000 Q D}{2 ϵ_{o}}$
$\Rightarrow \frac{500 Q D}{ϵ_{o}}$

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A parallel plate capacitor has rectangular plates each of area 20cm2 is separated by a distance D the potential difference between the two plates?

Area =2 cm2=2010000m2=21000m2Let distance =D, charge =QC=(ϵo)AD=ϵo2/100D↓CapacitancePD=V=Q/C=Q/ϵo×2/100D⇒1000QD2ϵo⇒500QDϵo

A parallel plate capacitor has rectangular plates each of area $20 c m^{2}$ is separated by a distance D the potential difference between the two plates?