Question

derive an expression for e lo electric field intensity in a parallel plate capacitor.
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Answer 1

$\huge{\underline{\tt{\blue{Derivation-}}}}$

Let the two plates are kept parallel to each other separated be a distance d and cross-sectional area of each plate is A.  

Electric field by a single thin plate

$E'=\dfrac {\sigma} {2E_o}$

$Total \ electric \ field \ between \ the \ plates$ $E = \dfrac{\sigma} {2E_o}+ \dfrac{\sigma} {2E_o}$

$or \ E= \dfrac{ \sigma} {E_o}$

$or \ E = \dfrac{Qd}{AE_o}$

$Potential \ difference \ between \ the \ plates$ $V=Ed$

$V = \dfrac {Qd}{AE_o}$

$Capacitance \ C= \dfrac {Q}{V}$

$Thus \ we \ get \ capacitance \ of \ parallel \ plate \ capacitor \ C= \dfrac {AE_o}{d}$