# 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

Homework Helper
Last edited:
• bluesteels
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?

No. In that region the expression for the field becomes more complicated (it becomes an integral, if you are familiar with those $$E = \int \frac Q {4\pi\varepsilon_0\, r^2}\ \,dy\, dx \quad$$.)
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