[5th repost] What is the probablity that electron will escape?

Question

One negative charge -Q is held in fixed position (x,y,z) = (1/4,1/4,1/4),
another charge is positive +2Q at (x,y,z) = (1,1,1).

What is the probablity that electron randomly placed
inside the cube 0

Answer 1

Okay, I'll put an end to this one. The answer you seek is
(π√(3)/16). You've carefully selected the coordinates of the -Q charge at (1/4,1/4,1/4) while the +2Q charge is at (1,1,1), so that the sphere of 0 electric potential would have its center at (0,0,0), while its radius would be (1/2)√(3). So, it's just a matter of working out the volume of the sphere within the (1,1,1) cube, and that's the probability that the electron will escape, because then it has a net positive potential necessary to escape the system. Numerically, it's about 34%.

Of course, I'm still having troubles convincing myself that any chaotic trajectory the electron enters will ALWAYS ultimately leave the system, even though "theoretically" it should.

Answer 2

The game here is to find the volume within the cube that is in the potential well (or hill) of the positive charge.

So find the potential everywhere using the usual method V = -kq/r on each charge and adding them up using the superposition principal.

Take the gradient of this function and set it to zero. This should give you a surface.

Pray to the gods of doable problems that the surface has some tractable form and you can find the volume around the positive charge enclosed by the surface and the cube.

Good luck.

Answer 3

answer in user-friendly electronics e book you will loose electrons thats referred to as a hollow fee yet a floater will come via and fill it up this all happens at 186,000 mile a sec so do no longer seek for it oh there is going one now