## Electric Potential Energy

An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of 5.43 * 10-11 m. What is EPEfinal - EPEinitial, which is the change in the electric potential energy?

I have no idea where to start
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 Quote by nckaytee An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of 5.43 * 10-11 m. What is EPEfinal - EPEinitial, which is the change in the electric potential energy? I have no idea where to start
Isn't the potential energy given by PE = kq1*q2/r

If EPE at infinity is 0, then ... The change is ...
 In my notes I have $$\Delta E.P.E$$ of electron = q(Vb-Va) = (-e)(Vb-Va) I am really not getting this Im sorry, I might have something here... One minute..

## Electric Potential Energy

Okay I am confused

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 Quote by nckaytee In my notes I have $$\Delta E.P.E$$ of electron = q(Vb-Va) = (-e)(Vb-Va) I am really not getting this Im sorry, I might have something here... One minute..
What is the charge of an electron and a proton?

 Quote by Wikipedia e = –1.602176487(40) × 10–19C p = 1.60217653(14)×10−19 C
 So, would it be... (-1.6*10^19)(0 - 5.43 *10^-11)

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 Quote by nckaytee So, would it be... (-1.6*10^19)(0 - 5.43 *10^-11)
It's the product of both charges.

PE = kq1*q2/r

And check the exponent of the charges. And don't forget k.
 Okay, so I did PE = [(9*10^9)(-1.6*10^-19)(1.6*10^-19)] / 5.43*10^-11 My answer is right!!! -4.24 * 10^18 Thank you

Recognitions:
Homework Help
 Quote by nckaytee Okay, so I did PE = [(9*10^9)(-1.6*10^-19)(1.6*10^-19)] / 5.43*10^-11 My answer is right!!! -4.24 * 10^18 Thank you
You're welcome.

And don't act so surprised. You had it nailed from kqq/r.

Good luck.

 Tags electric, energy, potential