How Does Charge Distribution Affect Electric Field Calculations?

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The discussion focuses on calculating the electric field and surface charge density for a uniformly charged square sheet and a conducting slab. The electric field just above the center of the nonconducting sheet is calculated using Gauss's law, but initial attempts yielded incorrect results. The surface charge density for the conducting slab is derived from the total charge divided by the total area, including all sides of the slab. Participants emphasize the importance of visualizing the problem and correctly accounting for the geometry of the charge distribution. Clarifications on the calculations and methods are sought to resolve discrepancies in the answers.
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Homework Statement


A charge of 4.1 nC is placed uniformly on a square sheet of nonconducting material, with side 19.0 cm, that lies in a horizontal plane.

1. What is the magnitude of the electric field just above the center of the sheet?

The same charge is placed on a horizontal square conducting slab of side 19.0 cm and vertical thickness 1.0 mm. Assume that the charge distributes itself uniformly on the large square surfaces.

2. What is the surface charge density?
3. What is the magnitude of the electric field just below the center of the slab?

Homework Equations


guass's law, Φ=EA

The Attempt at a Solution


1. use Φ=q/ε=EA so E = q/(εA) plug in numbers, but the answer was wrong.

2. σ = q / A I added up all sides of the slab, but the answer was wrong.

3. i did it the same way i did for 1, the answer was wrong.Would someone kindly tell me what was wrong with my calculation? thanks.
 
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## A = 0.19\cdot0.19~m^2 = 0.0361~m^2## right?
 
henxan said:
## A = 0.19\cdot0.19~m^2 = 0.0361~m^2## right?
Yes, that is how i calculated the area
 
So, doesn't the sheet have two sides?
 
henxan said:
So, doesn't the sheet have two sides?

thank you, Henxan!

The sheet does have two sides and therefore i got the correct answer for 1 :P

However, do you know how am i supposed to do the other two parts of the problem?
If you do, would you kindly help me. :)
 
I would strongly suggest that you draw your problems - actually mostly so you can visualize the problems better. :)..

On conducting surface, the charge distributes evenly on the entire surface, so ##\rho = \frac{q}{A_{tot}}##. If there is only a slight difference in the answer, try including the 4x1mmx19cm sides.
 
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