Gauss's Law and Charge Density
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Gauss's Law and Charge Density

[From: ] [author: ] [Date: 11-10-07] [Hit: ]
p = Q/vol = (8 * 10^-11)C / (0.=> E = p*d_z/εo = 0.......
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An insulating slab 5 mm x 2 m x 2 m has a charge of 8 x 10^-11 coulomb distributed uniformly throughout its volume. Determine the electric field at point P, which is located within the slab beneath its center, 1 mm from one of the faces.

The answer is 0.68 N/C but I do not know how to get there. Please explain each step to me thoroughly as I likely won't be able to understand what you do to solve the problem. Thank you in advance!

-
Place origin of coordinates at the center of the slab
and let the total flux pass though both faces (2A=2*d_x*d_y) of
Gaussian box (2*d_x*d_y*d_z) such that the electric flux is
parallel to the z-axis only.

By Gauss's law and symmetry:
Flux = E.2A = Q_enc/εo
Flux = E2A = p*Vol / εo
Flux = E2(d_x*d_y) = p*(d_x*d_y*2*d_z)/εo
=> E = p*d_z/εo

Data:
Let permittivity of slab be approximately:
εo=8.85*10^-12 F/m

Distance from origin to point in question:
d_z=0.0015 m

Uniform charge density of slab:
p = Q/vol = (8 * 10^-11)C / (0.005 * 2 * 2)m^3

=> E = p*d_z/εo = 0.68 N/c (with rounding)
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