Quick question about which radius to use on Gauss' law problem

In summary, for part (a) when looking up the equation E=Q/2e(0.8)^2, the value of 0.100mm is not used because it is close enough to the sheet to be considered an infinite sheet. For part (b), the value of 100m is used because it is far enough away that the sheet can be approximated as a point charge. If a different distance is given, the expression for the field will become more complicated.
  • #1
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Homework Statement
A square insulating sheet 80.0 cm on a side is held horizontally. The sheet has 4.50 nC of charge spread uniformly over its area. (a) Estimate the electric field at a point 0.100 mm above the center of the sheet. (b) Estimate the electric field at a point 100 m above the center of the sheet.
Relevant Equations
E= Q/2eA
Q= (4.5*10^-9)
confused on part A/B when I look up they did E= Q/2e(0.8)^2.

But why not use the 0.100mm because that is the area of the enclosed.

Same with B why did they use 100m and not 0.8m because 0.8 is smaller so it enclosed the charge
 
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  • #3
BvU said:
Hi,

For part (a) you are so close to the sheet that it's as good as an infinite sheet. So you get
$$E = \frac \sigma {2\varepsilon_0}\ .$$
For part (b) you are so far away that the sheet looks almost like a point charge, so you get
$$E = \frac Q {4\pi\varepsilon_0 r 2}\ .$$
##\ ##
Thank you now it makes sense. But for clarification, if they have given us 1m instead of 100m then we use 0.8m?
 
  • #4
No. In that region the expression for the field becomes more complicated (it becomes an integral, if you are familiar with those $$E = \int \frac Q {4\pi\varepsilon_0\, r^2}\ \,dy\, dx \quad $$.)

cf the disc of charge

##\ ##
 
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1. What is Gauss' law?

Gauss' law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is named after the German mathematician and physicist Carl Friedrich Gauss.

2. How do I determine which radius to use on a Gauss' law problem?

The radius used in a Gauss' law problem depends on the symmetry of the charge distribution. For a spherical charge distribution, the radius is the distance from the center of the sphere to the point where the electric field is being calculated. For a cylindrical or planar charge distribution, the radius is the distance from the axis of symmetry to the point where the electric field is being calculated.

3. Can I use any radius in a Gauss' law problem?

No, the radius used in a Gauss' law problem must be chosen carefully to ensure that the electric field is constant over the surface of the Gaussian surface. Using an incorrect radius can lead to incorrect results.

4. What if the charge distribution is not symmetrical?

If the charge distribution is not symmetrical, it may be necessary to divide the distribution into smaller, symmetrical parts and calculate the electric field for each part separately. The total electric field can then be found by summing the individual electric fields.

5. How is Gauss' law used in real-world applications?

Gauss' law is used in many real-world applications, such as calculating the electric field inside a capacitor, determining the electric field of a point charge, and analyzing the electric field of a charged sphere. It is also used in the design and analysis of electronic devices, such as computer chips and solar cells.

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