# Solving for Electrical Potential Energy: Durcell's Example

• cwill53
In summary: No, it is a double sum because we are considering pairs. For each charge j=r in the outer sum we add up all the potential energies to which it contributes as a member of a pair, so the inner sum runs over all those partners, i.e. all except r.But each pair (r,s) will get counted twice, once as j=r, k=s and again as j=s, k=r, so we halve the total.It might be clearer if the sum were written ##\Sigma_j\Sigma_{k>j}##, avoiding the double...
cwill53
Homework Statement
What is the potential energy of an arrangement of eight negative charges on the corners of a cube of side b, with a positive charge in the center of the cube? Suppose each negative charge
is an electron with charge −e, while the central particle carries a double positive
charge, 2e?
Relevant Equations
$$U=\frac{1}{2}\sum_{j=1}^{N}\sum_{k=j}^{}\frac{1}{4\pi \epsilon _{0}}\frac{q_{j}q_{k}}{r_{jk}}$$
This question is an example in Durcell's Electricity and Magnetism.
The solution goes as follows:
[In this case] there are four different types of pairs. One type involves the center charge, while the other three involve the various edges and diagonals of the cube. Summing over all pairs yields
$$U=\frac{1}{4\pi \epsilon _0}(8\cdot \frac{(-2e^2)}{(\frac{\sqrt{3}}{2})b}+12\cdot \frac{e^2}{b}+12\cdot \frac{e^2}{\sqrt{2}b}+4\cdot \frac{e^2}{\sqrt{3}b})$$
Then the book introduces the following equation to sum over pairs
$$U=\frac{1}{2}\sum_{j=1}^{N}\sum_{k=j}^{}\frac{1}{4\pi \epsilon _{0}}\frac{q_{j}q_{k}}{r_{jk}}$$

I'm just a bit confused at the meaning of finding the electrical potential energy. Can someone explain what these equations mean intuitively and then how to apply the equation to the above example? Thanks.

cwill53 said:
Homework Statement:: What is the potential energy of an arrangement of eight negative charges on the corners of a cube of side b, with a positive charge in the center of the cube? Suppose each negative charge
is an electron with charge −e, while the central particle carries a double positive
charge, 2e?
Relevant Equations:: $$U=\frac{1}{2}\sum_{j=1}^{N}\sum_{k=j}^{}\frac{1}{4\pi \epsilon _{0}}\frac{q_{j}q_{k}}{r_{jk}}$$

This question is an example in Durcell's Electricity and Magnetism.
The solution goes as follows:
[In this case] there are four different types of pairs. One type involves the center charge, while the other three involve the various edges and diagonals of the cube. Summing over all pairs yields
$$U=\frac{1}{4\pi \epsilon _0}(8\cdot \frac{(-2e^2)}{(\frac{\sqrt{3}}{2})b}+12\cdot \frac{e^2}{b}+12\cdot \frac{e^2}{\sqrt{2}b}+4\cdot \frac{e^2}{\sqrt{3}b})$$
Then the book introduces the following equation to sum over pairs
$$U=\sum_{j=1}^{N}\sum_{k=j}^{}\frac{1}{4\pi \epsilon _{0}}\frac{q_{j}q_{k}}{r_{jk}}$$

I'm just a bit confused at the meaning of finding the electrical potential energy. Can someone explain what these equations mean intuitively and then how to apply the equation to the above example? Thanks.
The process here is to figure how much energy must be put into assemble these charges. You can figure the total energy as the pairwise energy sum of all the constituent particles Each pair is counted only once and the total might be negative.
That being said the sum over pairs I would write as $$U=\sum_{j=1}^{N}\sum_{k=j+1}^{N}\frac{1}{4\pi \epsilon _{0}}\frac{q_{j}q_{k}}{r_{jk}}$$ Your method (I think) counts everything twice and divides by 2...
The book method just segregates the sum into sets of pairs the same distance apart for convenience in calculating. (So write out the terms in the sum and see the equivalence!).

Just to point out that the inner sum in the OP should be over ##k \neq j##, not ##k = j##.

cwill53
Orodruin said:
Just to point out that the inner sum in the OP should be over ##k \neq j##, not ##k = j##.
I was just getting ready to correct it, thanks. I still don’t understand what that part means.

cwill53 said:
I was just getting ready to correct it, thanks. I still don’t understand what that part means.
That the inner sum sums over all possible values of ##k## except ##j##. The double sum therefore sums over all possible combinations of ##j## and ##k## such that ##j## and ##k## are distinct. However, j=1 and k=2 represents the same pair of charges as j=2 and k=1 and so on. The potential energy of each pair is therefore counted twice, which is the reason for the factor of 1/2 in front.

Orodruin said:
That the inner sum sums over all possible values of ##k## except ##j##. The double sum therefore sums over all possible combinations of ##j## and ##k## such that ##j## and ##k## are distinct. However, j=1 and k=2 represents the same pair of charges as j=2 and k=1 and so on. The potential energy of each pair is therefore counted twice, which is the reason for the factor of 1/2 in front.
So is the inner sum as to avoid counting the pairs more than two times?

cwill53 said:
So is the inner sum as to avoid counting the pairs more than two times?
No, it is a double sum because we are considering pairs. For each charge j=r in the outer sum we add up all the potential energies to which it contributes as a member of a pair, so the inner sum runs over all those partners, i.e. all except r.
But each pair (r,s) will get counted twice, once as j=r, k=s and again as j=s, k=r, so we halve the total.
It might be clearer if the sum were written ##\Sigma_j\Sigma_{k>j}##, avoiding the double counting.

cwill53
haruspex said:
No, it is a double sum because we are considering pairs. For each charge j=r in the outer sum we add up all the potential energies to which it contributes as a member of a pair, so the inner sum runs over all those partners, i.e. all except r.
But each pair (r,s) will get counted twice, once as j=r, k=s and again as j=s, k=r, so we halve the total.
It might be clearer if the sum were written ##\Sigma_j\Sigma_{k>j}##, avoiding the double counting.
I’m sorry, but I’m really having trouble understanding how to apply this equation to this problem for some reason. I’ll write out an example of how I think I’m supposed do it, it would be great if you can correct where I go wrong.

haruspex said:
No, it is a double sum because we are considering pairs. For each charge j=r in the outer sum we add up all the potential energies to which it contributes as a member of a pair, so the inner sum runs over all those partners, i.e. all except r.
But each pair (r,s) will get counted twice, once as j=r, k=s and again as j=s, k=r, so we halve the total.
It might be clearer if the sum were written ##\Sigma_j\Sigma_{k>j}##, avoiding the double counting.
Is it that I do the following summation, taking j=1,2,...,N but for every time I take a summation of j=N, ##N\in ##{1,2,3,4,5,6,7,8,9} (for the nine charges including the charge of +2e), k is NOT equal to j? In this sense, would I be taking j and summing over k a total of 8 different times for every k, since k isn't equal to j? For the first set where j=1 I would have addend of U, being
$$U_1=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{1}q_{4}}{r_{14}}+\frac{q_{1}q_{5}}{r_{15}}+\frac{q_{1}q_{6}}{r_{16}}+\frac{q_{1}q_{7}}{r_{17}}+\frac{q_{1}q_{8}}{r_{18}}+\frac{q_{1}q_{9}}{r_{19}}))$$
Then the second addend would be
$$U_2=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{2}q_{1}}{r_{21}}+\frac{q_{2}q_{3}}{r_{23}}+\frac{q_{2}q_{4}}{r_{24}}+\frac{q_{2}q_{5}}{r_{25}}+\frac{q_{2}q_{6}}{r_{26}}+\frac{q_{2}q_{7}}{r_{27}}+\frac{q_{2}q_{8}}{r_{28}}+\frac{q_{2}q_{9}}{r_{29}}))$$
$$U_3=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{3}q_{1}}{r_{31}}+\frac{q_{3}q_{2}}{r_{32}}+\frac{q_{3}q_{4}}{r_{34}}+\frac{q_{3}q_{5}}{r_{35}}+\frac{q_{3}q_{6}}{r_{36}}+\frac{q_{3}q_{7}}{r_{37}}+\frac{q_{3}q_{8}}{r_{38}}+\frac{q_{3}q_{9}}{r_{39}}))$$
and so on, making

$$U_{TOT}=U_1+U_2+U_3+...+U_N, N=9$$

Is this correct?

Last edited:
cwill53 said:
Is it that I do the following summation, taking j=1,2,...,N but for every time I take a summation of j=N, ##N\in ##{1,2,3,4,5,6,7,8,9} (for the nine charges including the charge of +2e), k is NOT equal to j? In this sense, would I be taking j and summing over k a total of 8 different times for every k, since k isn't equal to j? For the first set where j=1 I would have addend of U, being
$$U_1=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{1}q_{4}}{r_{14}}+\frac{q_{1}q_{5}}{r_{15}}+\frac{q_{1}q_{6}}{r_{16}}+\frac{q_{1}q_{7}}{r_{17}}+\frac{q_{1}q_{8}}{r_{18}}+\frac{q_{1}q_{9}}{r_{19}}))$$
Then the second addend would be
$$U_2=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{2}q_{1}}{r_{21}}+\frac{q_{2}q_{3}}{r_{23}}+\frac{q_{2}q_{4}}{r_{24}}+\frac{q_{2}q_{5}}{r_{25}}+\frac{q_{2}q_{6}}{r_{26}}+\frac{q_{2}q_{7}}{r_{27}}+\frac{q_{2}q_{8}}{r_{28}}+\frac{q_{2}q_{9}}{r_{29}}))$$
$$U_3=\frac{1}{2}\frac{1}{4\pi \epsilon _0}((\frac{q_{3}q_{1}}{r_{31}}+\frac{q_{3}q_{2}}{r_{32}}+\frac{q_{3}q_{4}}{r_{34}}+\frac{q_{3}q_{5}}{r_{35}}+\frac{q_{3}q_{6}}{r_{36}}+\frac{q_{3}q_{7}}{r_{37}}+\frac{q_{3}q_{8}}{r_{38}}+\frac{q_{3}q_{9}}{r_{39}}))$$
and so on, making

$$U_{TOT}=U_1+U_2+U_3+...+U_N, N=9$$

Is this correct?
Yes, and note that you have ##q_1q_2## in one sum and ##q_2q_1## in another, hence the need to halve the total.

haruspex said:
Yes, and note that you have ##q_1q_2## in one sum and ##q_2q_1## in another, hence the need to halve the total.
My goodness, all that TeX I had to edit because I put in double subscripts initially for everything!

I see it now. Thanks a lot. I'm just glad I can understand it now.

hutchphd
cwill53 said:
My goodness, all that TeX I had to edit because I put in double subscripts initially for everything!
Which is why I recommended the "triangular" sum in post #2!...same result... .

cwill53

## 1. What is electrical potential energy?

Electrical potential energy is the energy stored in an electrical system due to the interaction between charged particles, such as electrons and protons.

## 2. How is electrical potential energy calculated?

Electrical potential energy is calculated by multiplying the charge of the particle by the electric potential at that point in the system.

## 3. What is Durcell's example in relation to solving for electrical potential energy?

Durcell's example is a hypothetical scenario often used in physics to demonstrate how to calculate electrical potential energy in a simple circuit.

## 4. What are the key equations used in solving for electrical potential energy in Durcell's example?

The key equations used in solving for electrical potential energy in Durcell's example are Ohm's Law (V=IR), the equation for electric potential (V=kQ/r), and the equation for electrical potential energy (U=qV).

## 5. How can I apply the concept of electrical potential energy in real-world situations?

Electrical potential energy is a fundamental concept in understanding and analyzing electrical systems, so it can be applied in various real-world situations such as designing circuits, calculating the energy consumption of electronic devices, and understanding the behavior of lightning strikes.

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