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

[bluesteels]

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

 

[BvU]

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
$$E = \frac Q {4\pi\varepsilon_0\; r^2}\ .$$
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[bluesteels]

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
$$E = \frac Q {4\pi\varepsilon_0 r 2}\ .$$
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Thank you now it makes sense. But for clarification, if they have given us 1m instead of 100m then we use 0.8m?

 

[BvU]

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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