Assuming a Parallel Plate Capacitor of area A, separation distance d, plate charges [tex]\pm Q[/tex]: and plate charge densities [tex]\pm\sigma[/tex]:(adsbygoogle = window.adsbygoogle || []).push({});

In my textbook and in Wikipedia,

Electric Field of a charged plane with large or infinite area: [tex]\frac{\sigma}{\epsilon}[/tex]

Electric Field between plates of parallel plate capacitor: [tex]\frac{\sigma}{\epsilon}[/tex]

Voltage difference between plates of parallel plate capacitor = [tex]\frac{Q \cdot d}{\epsilon \cdot A}[/tex]

My question is since a parallel plate capacitor contains two plates of equal and opposite charge, shouldn't there be an additional factor of two? In other words,

shouldn't the net electric field between the two plates be: [tex]\frac{2 \cdot \sigma}{\epsilon}[/tex]

and shouldn't the voltage difference between the two plates be: [tex]\frac{2 \cdot Q \cdot d}{\epsilon \cdot A}[/tex]

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# Simple Parallel Plate Capacitor Question

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