A) \[0\,v/m\]
B) \[\frac{\sigma }{2{{\varepsilon }_{0}}}v/m\]
C) \[\frac{\sigma }{{{\varepsilon }_{0}}}v/m\]
D) \[\frac{2\sigma }{{{\varepsilon }_{0}}}v/m\]
Correct Answer: C
Solution :
(c) \[\frac{\sigma }{{{\varepsilon }_{0}}}v/m\] From Gauss law, electric field due to charged infinite plane sheet, \[{{E}_{1}}=\frac{\sigma }{2{{\varepsilon }_{0}}}\]= Here, the field between the plates \[E={{E}_{1}}={{E}_{2}}\] \[=\frac{\sigma }{2{{\varepsilon }_{0}}}+\frac{\sigma }{2{{\varepsilon }_{0}}}=\frac{\sigma }{{{\varepsilon }_{0}}}v/m\]
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