0
$\begingroup$

If the term electrostatic potential energy is used to define the electric potential energy of the system and not an individiual charge in a closed system of two/more charged particles, how is it possible to find the electric potential energy of an individual charge using V= $\frac{U}{q}$ where V is the electric potential at a point due to the electric field, U is the electric potential energy and q is the magnitude of charge. Won't U in this case refer to the electric potential energy of the charge q?

$\endgroup$

2 Answers 2

1
$\begingroup$

Quite simply, because $V(r)$ is due to the other charge or charges in the system. You can't have a $V$ without a charge to create it.

The potential energy that you calculate using $U=qV$ is the potential energy of the system comprising the charges that create $V$, and the charge who's value is $q$.

$\endgroup$
6
  • 1
    $\begingroup$ Worth adding that the converse is sometimes helpful. To get the potential energy of the system, you add one charge at a time. The work done to add one more charge keeps getting harder - an intuitive way to explain that for some configurations potential energy has a $Q^2$ term... $\endgroup$
    – Floris
    Commented Sep 27, 2017 at 11:19
  • $\begingroup$ Just to clarify, if I had a system consisting of two particles of charge q1 and q2 and q1>>q2; if I did some work to move q2 away from q1 such that it no longer experiences force due to q1, would I be changing the potential energy of q2, if I use U=qV $\endgroup$
    – GreenApple
    Commented Sep 27, 2017 at 12:44
  • $\begingroup$ @GreenApple The potential energy is "stored" in the system ie both charges and so you cannot say that one charge has a certain amount of the potential energy and the other charge has another amount of potential energy. $\endgroup$
    – Farcher
    Commented Sep 27, 2017 at 13:35
  • $\begingroup$ But are there any cases where I can say that the electrostatic potential energy of only one charge has changed? ( example in a ball earth system we usually state that the ball's gravitational potential energy has changed, when the balls position relative to earth changes since the earth is so massive compared to the ball) $\endgroup$
    – GreenApple
    Commented Sep 27, 2017 at 14:31
  • $\begingroup$ We usually say that the ball's potential energy has changed, but that's shorthand. It's not correct, but as you point out, it is used and is convenient. I can tell you that novice students sometimes take it too seriously resulting in a poor idea of what potential energy is. And almost every textbook introduces potential energy in that incorrect way, so beginners' first introduction to the concept is flawed, and first impressions stick. But if the charges that create $V$ are fixed, you can loosely say that the PE of the charge changes. Example: parallel plate capacitor. $\endgroup$
    – garyp
    Commented Sep 27, 2017 at 16:52
0
$\begingroup$

AS @garyp has stated the potential (pun!) for confusion about potential energy can start early in ones education.
It is difficult enough to explain what potential energy is and that is perhaps why the extra layer of sophistication is omitted?

Consider the gravitational potential energy possessed by ball and Earth system.
If the Earth is not present the ball does not experience the force of gravitational attraction.

If equal magnitude and opposite direction forces are applied on the ball and the Earth to make their separation increase the work done by those forces increases the gravitational potential of the ball and Earth system.
If the ball is allowed to fall the gravitational potential of the ball and Earth system decreases and the kinetic energy of the ball and the Earth increases.
But what do you actually "observe"?
You observe the ball falling down towards the Earth but you do not observe the Earth moving up towards the ball although it has to to make sure that momentum of the ball and Earth system is conserved.

$M_{\rm Earth}\, v _{\rm Earth} + m_{\rm ball} V_{\rm ball} = 0$ where $M_{\rm Earth} \gg m_{\rm ball}$ hence $v _{\rm Earth} \ll V_{\rm ball}$.

So when you write

$\text{gpe lost by ball} = \text{kinetic energy gained by ball}$

you are really writing a shorthand version of

$\text{gpe lost by ball & Earth} = \text{kinetic energy gained by ball} + \text{kinetic energy gained by Earth} $

acknowledging the fact that the $\text{kinetic energy gained by Earth} $ is negligible compared with the $\text{kinetic energy gained by Earth}$.

This can also happen when you are doing electrostatics experiments where one of the charge carriers is "attached" to the Earth and the other charge carrier is much less massive and can be "observed" to move so you write

$\text{electric pe lost by electron} = \text{kinetic energy gained by electron}$

although the electron on its own cannot have have electric potential energy, there must be a least one other charge present.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.