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Answer: b
Explanation: The flux density of any field is independent of the position (point). D = σ/2 =2 X 10-6(-az)/2 = -10-6.
To find the electric flux density ( mathbf{D} ) due to a uniform surface charge density ( sigma ) located in the plane ( z = 2 ), we can use Gauss’s law.
The electric flux density ( mathbf{D} ) near an infinite plane sheet of charge is given by:
[
mathbf{D} = frac{sigma}{2} hat{n}
]
where ( hat{n} ) is the unit normal vector pointing away from the surface. The surface charge density ( sigma = 2 , mu C/m^2 = 2 times 10^{-6} , C/m^2 ).
The normal vector ( hat{n} ) points in the positive ( z )-direction since the sheet is placed at ( z = 2 ) and we are considering points above the sheet.
Calculating the flux density ( mathbf{D} ):
[
mathbf{D} = frac{2 times 10^{-6} , C/m^2}{2} hat{k} = 10^{-6} , C/m^2 hat{k}
]
Since the point ( P(1, 1, 1) ) is below the plane at ( z = 2 ), we should consider the contribution of the plane charge in the ( z = 1 ) direction (down