If the charge on parallel plates is pm Q and | vedclass

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Question

(A) For a parallel plate capacitor,the electric field due to one plate is $E_1 = \frac{\sigma}{2\epsilon_0}$,where $\sigma = \frac{Q}{A}$ is the surfa

Vedclass · Accepted Answer

$E = \frac{Q}{\epsilon_0 A}$

Vedclass · Answer

(A) For a parallel plate capacitor,the electric field due to one plate is $E_1 = \frac{\sigma}{2\epsilon_0}$,where $\sigma = \frac{Q}{A}$ is the surface charge density.Since both plates produce electric fields in the same direction between them,the total electric field $E$ is the sum of the fields from both plates.$E = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$.Substituting $\sigma = \frac{Q}{A}$,we get $E = \frac{Q}{\epsilon_0 A}$.

If the charge on parallel plates is $\pm Q$ and the area is $A$,write the formula for the electric field between the two plates.

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