How Does Charge Distribution Affect Electric Field Calculations?

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In summary, for a nonconducting sheet with a charge of 4.1 nC placed on a square with side 19.0 cm, the magnitude of the electric field just above the center of the sheet can be calculated using Φ=q/ε=EA. For a horizontal square conducting slab with the same charge and dimensions, the surface charge density can be found by using σ=q/A, and the magnitude of the electric field just below the center of the slab can be calculated using the same formula. It may be helpful to draw out the problem to visualize the solutions.
  • #1
chongkuan123
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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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  • #2
## A = 0.19\cdot0.19~m^2 = 0.0361~m^2## right?
 
  • #3
henxan said:
## A = 0.19\cdot0.19~m^2 = 0.0361~m^2## right?
Yes, that is how i calculated the area
 
  • #4
So, doesn't the sheet have two sides?
 
  • #5
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. :)
 
  • #6
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.
 

1. What is a uniformly charged sheet?

A uniformly charged sheet is a two-dimensional surface with a constant charge density, meaning that the amount of charge per unit area is the same throughout the entire sheet.

2. How is a uniformly charged sheet different from a point charge?

A point charge is a single, concentrated charge at a specific location, while a uniformly charged sheet has a distributed charge over its entire surface. This means that a point charge has a finite size, while a uniformly charged sheet has an infinite size.

3. How is the electric field calculated for a uniformly charged sheet?

The electric field for a uniformly charged sheet is calculated using the equation E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space.

4. What is the direction of the electric field for a uniformly charged sheet?

The electric field for a uniformly charged sheet is perpendicular to the surface of the sheet and points away from the sheet for a positive charge and towards the sheet for a negative charge.

5. What practical applications does a uniformly charged sheet have?

Uniformly charged sheets have many practical applications, such as in parallel plate capacitors, where they are used to create a uniform electric field between the plates. They are also used in electrophoresis, a technique used in biology and chemistry to separate and analyze charged particles. Additionally, uniformly charged sheets are used in electrostatic painting and photocopying.

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