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Is there a formula in electrostatics analogous to escape velocity?

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diagopod
#1
May31-10, 09:15 PM
P: 98
I was trying to understand the similarities and differences between electrostatics and standard Newtonian mechanics, in particular gravitational formulas. One thing I was curious about is escape velocity. For gravity I've learned that it's sqrt(2GM/R). But for electrostatics, suppose I was trying to determine the equivalent formula. For example, suppose one wanted to know the velocity a negatively charged rocket on the surface of a positively charged sphere would have to attain in order to escape the electrostatic force of that sphere. How would one go about that? At first I'd think it would be sqrt(2KeQ/ R), but that doesn't have units of velocity, so was wondering what the right formula would be? Thanks for any guidance.
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arunma
#2
May31-10, 09:54 PM
P: 908
Let's see what we can derive.

First you set the potential energy at R equal to the kinetic energy at infinity:

[tex]KE(r=\infty) = PE(r=R)[/tex]

Or,

[tex]\dfrac{1}{2}mv^2=k_e\dfrac{Qq}{R}[/tex] (assuming escape from a spherical charge)

[tex]v=\sqrt{2k_e\dfrac{q}{m}\dfrac{Q}{R}}[/tex]

So assuming I did the math right, it looks like your answer was correct, except that you neglected the charge to mass ratio of the object. The quantity q/m appears often in mechanics calculations involving electrostatic or magnetostatic forces. The reason it doesn't appear in the formula for gravity is because the mass of the projectile appears on both sides of the equation and will cancel out. What you're seeing is a direct result of the fact inertial and gravitation mass are equivalent, and that there's no parallel in electrodynamics. Pretty cool, huh?
diagopod
#3
Jun1-10, 12:40 AM
P: 98
Quote Quote by arunma View Post
Let's see what we can derive.

First you set the potential energy at R equal to the kinetic energy at infinity:

[tex]KE(r=\infty) = PE(r=R)[/tex]

Or,

[tex]\dfrac{1}{2}mv^2=k_e\dfrac{Qq}{R}[/tex] (assuming escape from a spherical charge)

[tex]v=\sqrt{2k_e\dfrac{q}{m}\dfrac{Q}{R}}[/tex]

So assuming I did the math right, it looks like your answer was correct, except that you neglected the charge to mass ratio of the object. The quantity q/m appears often in mechanics calculations involving electrostatic or magnetostatic forces. The reason it doesn't appear in the formula for gravity is because the mass of the projectile appears on both sides of the equation and will cancel out. What you're seeing is a direct result of the fact inertial and gravitation mass are equivalent, and that there's no parallel in electrodynamics. Pretty cool, huh?
Thanks! I really appreciate you walking me through this. That makes good sense all the way around, and yes, that's a remarkable parallel, the equivalence principle right, or no?

graphene
#4
Jun1-10, 01:46 AM
P: 196
Is there a formula in electrostatics analogous to escape velocity?

a little off, but you could see a similar example in an atom.
though in an atom we dont speak in terms of velocoties of electrons, we speak in terms of energies.
u must have heard of "ionization energy".
that's the energy difference between infinity (zero energy) and the energy of a shell(classically, if u know quantum mech, u wud say 'state')
diagopod
#5
Jun1-10, 04:35 PM
P: 98
Quote Quote by graphene View Post
a little off, but you could see a similar example in an atom.
though in an atom we dont speak in terms of velocoties of electrons, we speak in terms of energies.
u must have heard of "ionization energy".
that's the energy difference between infinity (zero energy) and the energy of a shell(classically, if u know quantum mech, u wud say 'state')
Thanks, so yes, I've heard the term, but not very strong on atomic physics. So are you saying that just as KE = PE for escape velocity in gravitation, the ionization energy (essentially KE) of the electron must equal the electrostatic potential energy in an atom in order for it to escape? That seems to have a lot of symmetry. If analogous to gravitation, then the equivalent of the "escape velocity" is twice the "orbital velocity" of the electron, which is essentially c x alpha at the bohr radius / ground state (I know, it's not really orbiting, but I think the approximate math holds right?). So the escape/ionization velocity would be 2c x alpha for that particular ground-state electron right? Or is it wrong to conclude that if the "orbital velocity" of the electron was doubled it would have enough energy to escape? Thanks again for your help.


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