Question

The intensity of an electric field between the plates of a charged condenser of plate area A will be :

• A. $A/\left(q{\epsilon }_0\right)$
• B. $qA/{\epsilon }_0$
• C. $q/\left({\epsilon }_{0}A\right)$
• D. none of these

Answer 1

The correct option is C $q/\left({\epsilon }_{0}A\right)$

Let the charge of the plate is $q$. using the Gauss's law, the electric field of a charge sheet is $E=\frac{q}{2A{ϵ}_0}$

now the electric filed in between the charged plate condenser is

$E^{\prime }=\frac{q}{2A{ϵ}_0}+\frac{q}{2A{ϵ}_0}=\frac{q}{A{ϵ}_0}$