Electric field intensity at points in between | vedclass

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Question

(B) The electric field intensity due to a single thin infinite sheet with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_

Vedclass · Accepted Answer

$0, \frac{\sigma}{\varepsilon_0}$

Vedclass · Answer

(B) The electric field intensity due to a single thin infinite sheet with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.For two parallel sheets with the same surface charge density $\sigma$:$1$. Between the sheets: The electric fields due to the two sheets are equal in magnitude but opposite in direction. Thus,the net electric field is $E_{net} = E_1 - E_2 = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$.$2$. Outside the sheets: The electric fields due to both sheets point in the same direction. Thus,the net electric field is $E_{net} = E_1 + E_2 = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$.Therefore,the electric field intensity between the sheets is $0$ and outside the sheets is $\frac{\sigma}{\varepsilon_0}$.The correct option is $B$.

Electric field intensity at points in between and outside two thin separated parallel sheets of infinite dimension with like charges of same surface charge density $(\sigma)$ are . . . . . . and . . . . . . respectively.

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