Answer:
According to Gauss' theorem, \[\oint{E\cdot dS=\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}}\] Since, angle between electric field and area vector is \[0{}^\circ .\] \[\therefore \] \[E\cdot dS=\frac{\sigma dS}{{{\varepsilon }_{0}}}\] \[(\because q=\sigma dS)\] \[E=\frac{\sigma }{{{\varepsilon }_{0}}}\Rightarrow E=\frac{\sigma }{{{\varepsilon }_{0}}}\hat{n}\] where, \[\hat{n}\] is unit vector in normal direction. So, electric field at point P due to each part \[=\frac{1}{2}E=\frac{\sigma }{2{{\varepsilon }_{0}}}\hat{n}\]
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