The given figure shows two parallel plates A | vedclass

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(D) The electric field due to an infinite charged sheet with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\epsilon_0}$.For plate $A

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Both $I$ and $III$

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(D) The electric field due to an infinite charged sheet with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\epsilon_0}$.For plate $A$ (charge density $+\sigma$),the electric field points away from the plate. For plate $B$ (charge density $-\sigma$),the electric field points towards the plate.In region $I$ (to the left of plate $A$): The field due to $A$ is $E_A = \frac{\sigma}{2\epsilon_0}$ (to the left) and the field due to $B$ is $E_B = \frac{\sigma}{2\epsilon_0}$ (to the right). The net electric field is $E_{net} = E_B - E_A = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$.In region $II$ (between the plates): The field due to $A$ is $E_A = \frac{\sigma}{2\epsilon_0}$ (to the right) and the field due to $B$ is $E_B = \frac{\sigma}{2\epsilon_0}$ (to the right). The net electric field is $E_{net} = E_A + E_B = \frac{\sigma}{\epsilon_0} 
eq 0$.In region $III$ (to the right of plate $B$): The field due to $A$ is $E_A = \frac{\sigma}{2\epsilon_0}$ (to the right) and the field due to $B$ is $E_B = \frac{\sigma}{2\epsilon_0}$ (to the left). The net electric field is $E_{net} = E_A - E_B = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$.Therefore,the electric intensity is zero in both regions $I$ and $III$.

The given figure shows two parallel plates $A$ and $B$ with charge densities $+\sigma$ and $-\sigma$ respectively. The electric intensity will be zero in which region?

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