Why Potential Energy cannot be included twice?

[CGandC]

Homework Statement

Why the total potential energy of a system of 2 charged particles q_1 , q_2 with distance 'r' between them is $$ k\frac{q_1 q_2}{r} $$ and not $$ 2k\frac{q_1 q_2}{r} $$?

Relevant Equations

$$ V = k\frac{ q_1}{r} $$
$$ U = q_2*V $$

If for example I have two charged particles q_1 , q_2 with distance 'r' between them, then:

The potential energy that results from particle q_1 exerting force on particle q_2 is $$ k\frac{q_1 q_2}{r} $$

If I do the same process for particle q_2:
The potential energy that results from particle q_2 exerting force on particle q_1 is $$ k\frac{q_1 q_2}{r} $$

Therefore, the total potential energy of the whole system is:
$$ U = k\frac{q_1 q_2}{r} + k\frac{q_1 q_2}{r} = 2k\frac{q_1 q_2}{r} $$

But it turns out that this result is false because I need to divide it by 2.
So my question is: why is this result false? why do I need to divide it by 2? why can't I just keep it as it is?

 

[gneill]

The potential energy is related to the energy required to assemble the system, not simply the forces acting on individual particles. When you put the first particle in place all by itself, no energy is required as there is no other electric field acting on it. Then, you bring the second particle into position (from an infinite distance), and as you do so it feels the force due to the charge of the first particle. That force is given by:

$F = k \frac{q_1 q_2}{R^2}$

The work done in bringing the second particle into position is equal in magnitude to the potential energy of the system.

If you integrate:

$W = \int_\infty^r k\frac{q_1 q_2}{R^2}\, dR$

you should get the expected result.

 

[DaveC426913]

Looking at it another way, you could include the potential energy for each of the two particles, but you then must bear in mind that - since the net result is that they'll meet somewhere in the middle - the r for each will be correspondingly reduced.

 

[CGandC]

I understand now, Thanks!

 

[PeroK]

The potential energy is related to the energy required to assemble the system, not simply the forces acting on individual particles. When you put the first particle in place all by itself, no energy is required as there is no other electric field acting on it. Then, you bring the second particle into position (from an infinite distance), and as you do so it feels the force due to the charge of the first particle. That force is given by:

$F = k \frac{q_1 q_2}{R^2}$

The work done in bringing the second particle into position is equal in magnitude to the potential energy of the system.

If you integrate:

$W = \int_\infty^r k\frac{q_1 q_2}{R^2}\, dR$

you should get the expected result.

Just to add to this. When I first heard this argument, I thought wait a minute: when you bring in $q_2$ then $q_1$ will move! It won't stay still while you try to assemble the system.

Then I realized that you can hold $q_1$ in place with another "restraining" force. And, as $q_1$ does not move this restraining force does no work, so does not change the energy of the system.