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(D) The electric field due to an infinite sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.For a s

Vedclass · Accepted Answer

both $A$ and $C$ are correct.

Vedclass · Answer

(D) The electric field due to an infinite sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.For a sheet with charge density $+\sigma$,the field is directed away from the sheet. For a sheet with charge density $-\sigma$,the field is directed towards the sheet.Outside the plates,the fields due to the two sheets are equal in magnitude and opposite in direction,so the net electric field is $E = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$.Between the plates,the fields due to both sheets are directed from the positive sheet to the negative sheet (towards the right). Thus,the net electric field is $E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$.Therefore,the statements in $A$ and $C$ are correct.

Two infinite sheets of uniform charge density $+\sigma$ and $-\sigma$ are parallel to each other as shown in the figure. The electric field is:

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