Electrical Potential Energy equations

[brood910]

TBR gives two equations for this, and I am wondering they are the same thing or not..

Change in PEq = kq1q2(1/r - 1/r)

Another one is = Change in PEq = qV

I know the second one well but I've never heard of the first one..
Can I set them equal to each other then?

Also, TBR says when attractive force is present, the PE will decrease when it explains about the first equation... If this is true, then change in PE would be a negative value.. BUT, doesnt attractive force makes final r smaller and thus makes change in PE positive?

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[polyatomic]

These two equations apply to two different situations.

Change in PEq = kq1q2 / r
This equation is for when you have two charges (hence the q1 and q2). And you measure the "potential energy" caused by the presence of one on the other.

I don't know if you know the force equation that's similar...only difference is radius is squared
F = kq1q2 / r2
This tells you the attractive force between two charges, dependent on the distance between them. So this could be a situation where you just place two charges next to one another. There does not need to be any external field.


Change in PEq = qV
This tells you the potential of a charge in the presence of an electric potential. So this would be used in a situation where you have two plates that generate an external voltage, and you place a charge around there.

 

[brood910]

These two equations apply to two different situations.

Change in PEq = kq1q2 / r
This equation is for when you have two charges (hence the q1 and q2). And you measure the "potential energy" caused by the presence of one on the other.

I don't know if you know the force equation that's similar...only difference is radius is squared
F = kq1q2 / r2
This tells you the attractive force between two charges, dependent on the distance between them. So this could be a situation where you just place two charges next to one another. There does not need to be any external field.


Change in PEq = qV
This tells you the potential of a charge in the presence of an electric potential. So this would be used in a situation where you have two plates that generate an external voltage, and you place a charge around there.

Apparently, you dont know this either... The first equation is kqq(1/rfinal - 1/rinitial)..
Thanks though..

 

[polyatomic]

Apparently, you dont know this either... The first equation is kqq(1/rfinal - 1/rinitial)..
Thanks though..

Oh. I guess that's for moving charges with respect to one another. Google: http://webpages.uah.edu/~bonamem/PH112_Spring07/chapter24.pdf

 

[brood910]

Oh. I guess that's for moving charges with respect to one another. Google: http://webpages.uah.edu/~bonamem/PH112_Spring07/chapter24.pdf

Nope... not moving them with respect to each other..
It is for either attractive or repulsive forces between them. I think this is not that important I guess since I cant find this in other MCAT books..

EDIT: NVM. PE = kqq/r. So Change in PE = kqq(1/rf - 1/ri).

 

[dudewheresmymd]

TBR gives two equations for this, and I am wondering they are the same thing or not..

Change in PEq = kq1q2(1/r - 1/r)

Another one is = Change in PEq = qV

I know the second one well but I've never heard of the first one..
Can I set them equal to each other then?

Also, TBR says when attractive force is present, the PE will decrease when it explains about the first equation... If this is true, then change in PE would be a negative value.. BUT, doesnt attractive force makes final r smaller and thus makes change in PE positive?

Take two unlike charges, (q1*Q2) will be negative, and rfinal < r initial (because they are moving towards one another), let's say r initial = infinity, therefore (negative product from q1*q2) *(positive 1/rf) = a negative, therefore the change in PE is negative if you place two unlike charges separated initially by a distance of infinity. Things that have a negative potential energy are MORE stable than things with a higher potential energy (analogous to grav potential energy) U=-Gm1m2/r (the farther away a planet is from the earth, the LESS negative (more positive) its grav potential energy, thus the "less stable" it is...same with grav potential energy on earth, mgh, the higher up a cliff the "less stable."

Attractive force F=kq1*q2/r^2, for two negative charges simply measures the force between the two charges in Newtons. It does not tell you anything about the change in potential energy, however as you mentioned if you were to multiply the attractive force * r, you get U=kq1q2/r which is then used to derive the potential energy formula as you mentioned.