Plate conductor in electrostatic equilibrium

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SUMMARY

The discussion focuses on the electrostatic properties of a thin, square conducting plate measuring 50.0 cm per side, carrying a total charge of 4.00 x 10-8 C. The charge density on each face of the plate is calculated to be 8 x 10-8 C/m2. Using Gauss's law, the electric field just above the plate is determined to be 9.04 kN/C, while the electric field just below the plate is -9.04 kN/C, indicating the direction of the electric field relative to the Gaussian surface. The negative sign in the electric field below the plate reflects the downward direction of the field, consistent with the orientation of the area vector.

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Homework Statement


A thin, square, conducting plate 50.0 cm on a side lies
in the xy plane. A total charge of 4.00 x 10-8 C is placed
on the plate. Find (a) the charge density on each face of
the plate, (b) the electric field just above the plate, and
(c) the electric field just below the plate. You may assume
the charge density is uniform.

Homework Equations



σ = Q/A

The Attempt at a Solution



a) σ = 4×10^-8 / (.5)^2 / 2 = 8x10^-8 C/m^2
b) Ok using gauss law

E = σ / ε = 9.04 kN/C

c) my problem is why for part C, the answer is -9.04 kN/C. If my Gaussian surface is below the plane, its area vector is pointing downward, and my positive bottom surface charges electric field is also pointing down, so they are pointing in the same direction, thus E should be positive??
 
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Woopydalan said:
c) my problem is why for part C, the answer is -9.04 kN/C. If my Gaussian surface is below the plane, its area vector is pointing downward, and my positive bottom surface charges electric field is also pointing down, so they are pointing in the same direction, thus E should be positive??

What you should conclude is that the flux is positive for your Gaussian surface. The field is downward (as you stated) and I think that is all that is meant by the answer being negative.
 

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