# How to find potential *Energy* between two point charges?

1. Oct 8, 2015

### gracy

How to find potential ENERGY between two point charges?I know how to find potential difference between two points and I also know how to find potential at any point .
potential difference between two points 1 and 2 is simply $V1$-$V2$
potential at any point r=V
While source charge is Q
$V(r)$=$Q/4πε0r$

2. Oct 8, 2015

### nasu

You have almost all you need to find out.
What is the PE of a charge q when located at a point with potential V, produced by charge Q?

3. Oct 8, 2015

### ehild

The formula is wrong.

4. Oct 8, 2015

### gracy

qV
i.e
$qQ$/$4(πε0r)$

5. Oct 9, 2015

### nasu

Yes. Just move the first paranthesis before the "4".
You don't have this in the textbook?

6. Oct 9, 2015

### gracy

But it is potential energy at a point .My textbook does not specify at which point between the particles,it just says determine the potential energy between the two particles.

7. Oct 9, 2015

### nasu

No, r is the distance between particles. There is no point involved. This is the PE of the system of two charges.

8. Oct 9, 2015

### gracy

$qQ/(4πε0r)$
Is it correct?

9. Oct 9, 2015

### gracy

$qQ/(4πε0r)$ is the potential energy between the two particles having charge q and Q respectively,it is also equal to potential energy of the any of the two particles due to the other one ,right?

10. Oct 9, 2015

### nasu

The potential energy is a property of the system not of an individual component. You can say "the pe of the system of two particles" or "the pe of one in the field of the other". In the second case one of them is "represented" by its field. But you cannot "split" it between the two.

11. Oct 9, 2015

### gracy

That's what I am saying.

12. Oct 9, 2015

### BvU

$\#\#$ V(r) = {Q\over 4\pi\epsilon_0 r} $\# \#$

$V(r) = {Q\over 4\pi\epsilon_0 r}$ is the electrical potential at a distance $r$ from the origin if a charge $Q$ is placed at the origin. $V$ is in volt (= Joule/Coulomb) .

It takes an energy $q V$ to bring a charge from a place where $V = 0$ (usually infinity) to a place where the potential energy is $V$. $\ \ qV$ is in Joule

So $qQ\over 4\pi\epsilon_0 r$ is the energy it takes to move $q$ from infinity to a distance $r$ of $Q$ . It is also the energy it takes to move $Q$ from infinity to a distance $r$ of $q$ .

13. Oct 9, 2015

### gracy

$V(r) = {Q\over 4\pi\epsilon_0 r}$
is potential energy of q when it is kept at distance r of Q from infinity.
It is also potential energy of Q when it is kept at origin (from infinity to a distance r of q)
It is also potential energy between the two particles having charge q and Q,Right?

14. Oct 9, 2015

### BvU

No. $$V(r) = {Q\over 4\pi\epsilon_0 r}$$is the electrical potential at a distance $r$ from the origin if a charge $Q$ is placed at the origin.

No q involved. $V$ is a function of position (or rather, distance) $r$.

15. Oct 9, 2015

### gracy

Sorry,sorry!I meant
${Qq\over 4\pi\epsilon_0 r}$
is potential energy of q when it is kept at distance r of Q from infinity.
It is also potential energy of Q when it is kept at origin (from infinity to a distance r of q)
It is also potential energy between the two particles having charge q and Q,Right?

16. Oct 12, 2015

### haruspex

Nearly right, but you are mixing two different statements.
You can think of it is the PE of Q when held at distance r from q (though as has been pointed out, it is really the PE of the system);
It is the energy required to bring a charge Q from infinity to be at distance r from q.

Writing " ... when kept at ... from infinity" makes it incomprehensible.

17. Oct 13, 2015

### gracy

18. Oct 13, 2015

Yes.