3 ms·

Yes, I actually kind of get the infinite part, but thanks for the explanation it's interesting. I was wondering what the intensity looks like for a distance fie

by tomxor 4y ago

Yes, I actually kind of get the infinite part, but thanks for the explanation it's interesting. I was wondering what the intensity looks like for a distance field around a finite plane and how that relates. This might be harder to describe. What I'm suspecting is that under a certain distance for a given finite plane it's almost constant similar to the infinite version, but outside of that distance there must be some non uniform falloff function.

penteract 4y ago
For a finite circle, where you're on a line perpendicular to it going through its center, the gravitational force you experience towards it is exactly proportional to the amount of your field of view it occupies - if I'm right, that should be proportional to `1-cos(atan(r/d))` where r is the radius of the circle and d is the distance between you and the center of the circle.
gus_massa 4y ago
You are correct. For a short distance it's a almost a constant 1 like an infinite wall with the same charge density. For a long distance is almost 1/r^2 like punctual source. For an intermediate distance, it's a mess and you hope nobody ask for an exact solution. A rectangle is a mess. It's easier with a circle. The approximations at short and long distance are the same, but a circle has an "easy" formula in between. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.h... 1-x/sqrt(x^2+R^2) where R is the radio of the circle. Let's use R=10 to keep it simple. For a short distance 0<x<R/10, then it's almost a constant https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%5E2%29++and+0+for+x+from+0+to+1 https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%... For a long distance, x>4R it's almost like R^2/(2*x^2) https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%5E2%29+and+10%5E2%2F%282x%5E2%29+for+x+from+0+to+100&lang=es https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%... There are some trick to add more term to the approximations to reduce the middle part where both approximations are bad, like using A+Bx or A+Bx+Cx^3 for short distance and D/x^2+E/x^4 or D/x^2+E/x^4+F/x^4 for long distance. It depends on how much you care about the precision and how many calculations you want to do. In some particular cases like a circle or the shell of a sphere there are closed formulas for the intermediate distances. In other cases there is no nice formula.

by tomxor 4y ago

penteract 4y ago

For a finite circle, where you're on a line perpendicular to it going through its center, the gravitational force you experience towards it is exactly proportional to the amount of your field of view it occupies - if I'm right, that should be proportional to `1-cos(atan(r/d))` where r is the radius of the circle and d is the distance between you and the center of the circle.

gus_massa 4y ago

You are correct. For a short distance it's a almost a constant 1 like an infinite wall with the same charge density. For a long distance is almost 1/r^2 like punctual source. For an intermediate distance, it's a mess and you hope nobody ask for an exact solution. A rectangle is a mess. It's easier with a circle. The approximations at short and long distance are the same, but a circle has an "easy" formula in between. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.h... 1-x/sqrt(x^2+R^2) where R is the radio of the circle. Let's use R=10 to keep it simple. For a short distance 0<x<R/10, then it's almost a constant https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%5E2%29++and+0+for+x+from+0+to+1 https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%... For a long distance, x>4R it's almost like R^2/(2*x^2) https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%5E2%29+and+10%5E2%2F%282x%5E2%29+for+x+from+0+to+100&lang=es https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%... There are some trick to add more term to the approximations to reduce the middle part where both approximations are bad, like using A+Bx or A+Bx+Cx^3 for short distance and D/x^2+E/x^4 or D/x^2+E/x^4+F/x^4 for long distance. It depends on how much you care about the precision and how many calculations you want to do. In some particular cases like a circle or the shell of a sphere there are closed formulas for the intermediate distances. In other cases there is no nice formula.