# Expression for Electric ptential energy for System of n' charges

## Recommended Posts

How to Derive an Expression for Electric ptential energy for System of n' charges?

##### Share on other sites

Superposition.

Note the field is conservative.

The potential V is therfore given by summing the individual contributions over all n.

$V = \frac{1}{{4\pi \varepsilon }}\sum\limits_1^n {\frac{{{q_i}}}{{{r_i}}}}$

Edited by studiot

##### Share on other sites

Superposition.

Note the field is conservative.

The potential V is therfore given by summing the individual contributions over all n.

$V = \frac{1}{{4\pi \varepsilon }}\sum\limits_1^n {\frac{{{q_i}}}{{{r_i}}}}$

All charges in the system will interact with each other. The correct expression would therefore be

$V = \frac{1}{4\pi \varepsilon} \sum\limits_{i < j} \frac{q_i q_j}{r_{ij}}$

where $r_{ij}$ is the distance between charge i and charge j. The sum is over all pairs of charges.

##### Share on other sites

All charges in the system will interact with each other.

I don't think so, that would be contrary to superposition.

Electric potential is an additive scalar that obeys superposition.

Are you sure you are not thinking of electric flux?

##### Share on other sites

It *is* superposition. Every charge will feel the electric potential of all the others, which is easy to check. I am 100% positive on this one,

Although it strikes mr now that you may be thinking of the potential of a test charge introduced into the system (although lacking the charge of the test charge). The potential energy of the entire system is given by the expression I wrote down.

In fact, it is easy to derive the full expression from your expression. Start with one charge, potential zero. For every charge you add, the potential energy is changed by an amount given by your formula (corrected with the charge of the new particle) with the sum going to j-1 when you introduce the jth charge. Hence the sum over i<j.

Edited by Orodruin

##### Share on other sites

Well, for the benefit of janaki, who asked this question, would you agree with me that my formula is correct if we consider the test point for the potential to be at the origin, and placed charge q at distance +r along the x axis?

sorry it's nearly 1am here so I can't wait any longer.

Look here you will see my formula

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html

Good night all.

Edited by studiot

##### Share on other sites

Well, for the benefit of janaki, who asked this question, would you agree with me that my formula is correct if we consider the test point for the potential to be at the origin, and placed charge q at distance +r along the x axis?

She asked for the potential energy of the system. Not the potential of a test charge. The potential energy of the system I can only interpret in one way. I agree that your formula with the sum multiplied by the test charge charge is the potential energy added by a test charge in the origin.

Look here you will see my formula

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html

Good night all.

Again, what is shown is the electric potential (measured, e.g., in Volts). What was asked for was the electrical potential energy (measured in, e.g., Joules) of the system. See http://en.wikipedia.org/wiki/Electric_potential_energy

Edited by Orodruin

##### Share on other sites

This is not right. and I am not thinking about test charges.

A single charge has a defined potential around it, with spherical symmetry, of potential inversely proportional to the distance from the charge.

However, re-reading the OP you are correct, she did ask for electric potential energy.

This is indeed different from the electric potential, which is measured in volts, not joules. So this is a good point.

And yes, to assess the electric potential energy you bring each charge in turn from infinity to its location and consider the work you do for each added charge.

So yes introducing the first charge will take zero work,. Work will be done against the electric field of this first charge when adding the second. This work may be positive or negative depending upon the signs of the charges. The third charge added will require work against the superposed fields of the first two and so on.

It is important to note that you need to modify your formula to a double summation to achieve this.

Also is is usual to change the 1/4 to 1/8 to offset the effect of counting each pair twice, although I see you have used the condition i<j instead.

##### Share on other sites

This is not right. and I am not thinking about test charges.

It is important to note that you need to modify your formula to a double summation to achieve this.

I could have been more clear. When I just say "potential" I mean potential energy.

To drop the second sum sign by lazyness when intention is clear (and I did indicate in the text that summation was over particle pairs) is standard in my field and several others I am aware of so this is in my backbone, as is dropping additional integral signs. When it comes to 1/2 vs i<j I find the latter more illustrative.

##### Share on other sites

Thankas all, How to find U?

##### Share on other sites

Hello janaki, I see you are a Medical Doctor. Congratulations. That is difficult.

Back to the matter in hand, the PE of a system of N charges.

I am not sure how much Physics you know or quite what your interest is in this calculation.

Do you, for instance know how to calculate the PE for a system of two charges only?

Would you like me to expand on the calculation procedure I outlined in post#8, perhaps from 'first principles'?

Edited by studiot

##### Share on other sites

All charges in the system will interact with each other. The correct expression would therefore be

$V = \frac{1}{4\pi \varepsilon} \sum\limits_{i < j} \frac{q_i q_j}{r_{ij}}$

where $r_{ij}$ is the distance between charge i and charge j. The sum is over all pairs of charges.

Analyze units..

V, electric potential, is in Volts, or Coulombs/Farads

http://en.wikipedia.org/wiki/Electric_potential

If you have Q * Q in numerator, you will have Coulombs ^2 in units.

$\frac{C^2}{F} \neq \frac{C}{F}$

When I just say "potential" I mean potential energy.

Electric potential energy has symbol UE (not V).

http://en.wikipedia.org/wiki/Electric_potential_energy

Thankas all, How to find U?

##### Share on other sites

Analyze units..

V, electric potential, is in Volts, or Coulombs/Farads

http://en.wikipedia.org/wiki/Electric_potential

If you have Q * Q in numerator, you will have Coulombs ^2 in units.

$\frac{C^2}{F} \neq \frac{C}{F}$

Electric potential energy has symbol UE (not V).

http://en.wikipedia.org/wiki/Electric_potential_energy

Since when does physics depend on the symbols used to represent it? The representation of the same quantity may be dependent on the field or even the specific paper you read. Potential energy will typically be denoted U, V, or EP and who knows what else may be used by people.

The question was for the potential energy of the system, I chose to represent it using a V, which is something I am free to do as long as it is clear from context (answering the OP).

Thanks all..

- swapna

## Create an account

Register a new account