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Old Feb 5th 2019, 03:29 PM   #1
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Made up a problem; did I screw up?

Been a long time since I did this stuff, so my error could be math or physics based.

Problem: You have a massless charge of q = +1 nC fixed in place. A small object with charge of +1 nC and mass of m = 10^-3 kg is dropped from a position 1 m above the massless charge. How close do the two come to touching? Assume the only gravity on the falling object is from Earth.

I'm using g = 10 m/s^2 (didn't want to break out calculator) and k = 9*10^9 N*m^2/C^2.

My first thought to simplify was that I could solve for when the work done by gravity equals the work done by the electric force as that would mean no change in kinetic energy, which means the speed was again 0 (same as starting). That would be the closest they get before the massless particle's repulsion finally overcomes the falling.

So I defined the distance the object falls as x and set the work of the electric force as the integral from 0 to x of kq^2/(1-x)^2 dx and the work of gravity as the integral from 0 to x of mg dx.

For the first integral, I got xkq^2/(1-x). For the second, I got (10^-2)x.

Then I set those equal to each other, and came out with x = 1 - (9*10^-7) meters, meaning it falls almost the entire distance.

This is only meant to account for basic stuff, so no need to get into the impossibility due to this being smaller than an object with that much mass could be or anything like that.

Thanks in advance for anyone willing to check this out.
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Old Feb 5th 2019, 08:20 PM   #2
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You mean something like this?

My guess is you add up the forces (gravity and electric) and apply newton's second law F=ma with the resultant force.
Burn those raisin muffins. Burn 'em all I say.
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Old Feb 6th 2019, 02:53 AM   #3
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kiwiheretic's method would require integrating the forces to get the velocities, and then integrating the velocities to get the displacements.
It could be done but...

The method put forward by LemonLizard (nice name by the way) seems sensible,
But I don't have the time (or the inclination) at the moment to check in detail.

I'm sure one of the more mathematically adept of our members will happily attack this little problem.
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