How Does Charge Distribution Affect Electric Field Calculations?

[chongkuan123]

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

Gauss'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.

 

[henxan]

$A = 0.19\cdot0.19~m^2 = 0.0361~m^2$ right?

 

[chongkuan123]

$A = 0.19\cdot0.19~m^2 = 0.0361~m^2$ right?

Yes, that is how i calculated the area

 

[henxan]

So, doesn't the sheet have two sides?

 

[chongkuan123]

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. :)

 

[henxan]

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.