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Question

(B) The electric field due to an infinite plane sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_{0}}$

Vedclass · Accepted Answer

$\sigma / \varepsilon_{0}$ towards the negatively charged plane

Vedclass · Answer

(B) The electric field due to an infinite plane sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_{0}}$.For the positively charged plane $(+\sigma)$,the electric field $E_{+}$ points away from the plane.For the negatively charged plane $(-\sigma)$,the electric field $E_{-}$ points towards the plane.Between the two planes,both fields point in the same direction,i.e.,from the positive plane to the negative plane.Therefore,the net electric field $E_{net} = E_{+} + E_{-} = \frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}} = \frac{\sigma}{\varepsilon_{0}}$.The direction is from the positive plane towards the negative plane.

Two infinite parallel metal planes contain electric charges with charge densities $+\sigma$ and $-\sigma$ respectively,and they are separated by a small distance in air. If the permittivity of air is $\varepsilon_{0}$,then the magnitude of the field between the two planes with its direction will be:

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