From a homework assignment, there are 4 spheres spaced 1cm apart. Each of the spheres are charged to +10nC and weigh 1 gram. The question wants us to find the final speed of the charges once they've drifted far apart.

4 charged spheres

I've found the answer, and I realized where my calculations were incorrect. What I don't understand is why my calculations are incorrect. To find the solution to this question, I realized I'd need to find the potential energy of the entire system. I assembled the potential energy with the following formula: $$ \Delta U=\int -F_edr = \frac{kq_1q_2}{r}$$ Which lead me to this incorrect equation: $$U_f-U_i=0-4(\frac{kQ}{L} + \frac{kQ}{L} + \frac{kQ}{\sqrt2 L})$$ The intuition behind this formula was the interaction of three spheres on one sphere would be $\frac{kQ}{L} + \frac{kQ}{L} + \frac{kQ}{\sqrt2 L}$ so the interaction of all spheres on each other would be 4 times that quantity. However, it turns out there are only 6 total interactions in the entire system: $$U_f - U_i = 0 - (\frac{kQ}{L} + \frac{kQ}{L} + \frac{kQ}{L} + \frac{kQ}{L} + \frac{kQ}{\sqrt2 L} + \frac{kQ}{\sqrt2 L})$$ But wouldn't this count interactions going in only one direction? Specifically, if the spheres were named A, B, C, and D––order doesn't matter. Then the only interactions that could have been noted would be $A\,\to\,B, A\,\to\,C, A\,\to\,D, B\,\to\,C, C\,\to\,D, B\,\to\,D$. But what about the half of the interactions: $B\,\to\,A, C\,\to\,A, D\,\to\,A, C\,\to\,B, etc.$ Does the other half not matter in the context of potential energies?


No, if you included the other ones you would be double-counting (notice your way has 12 terms instead of 6). An interaction involves two charges. So you have the A-B interaction, the A-C interaction, etc. The X-Y interaction Isn't the combination "X on Y" and "Y on X" interaction. It's just a single interaction between the two charges.

Another way to look at it is to use the other (but technically the same) definition of potential energy as the energy needed to construct the charge configuration by bringing in each charge slowly from infinity.

1) Bring charge $D$ in first for free.

2) Bring in charge $C$ which feels a force from $D$.

3) Bring in charge $B$ which feels a force from both $C$ and $D$.

4) Bring in charge $A$ which feels a force from charges $B$, $C$, and $D$.

And there are your six terms you need for the total energy.


The Question is more of a combinatorics question.

"What are the number of interactions between 6 particles;Given that each interaction involves 2 particles?" The answer is of course $4C2$ which is 6. (The Mistake You had done is that You had not considered an interaction Between 2 Particles at a time)

Now for the Physics Part. If the Particles are A,B C,D There are

  1. 4 interaction between adjacent particles (A,B),(B,C),(C,D),(D,A) Giving PE $4(kQ*Q/L)$
  2. 2 interactions between "opposite" particles (A,C),(B,D) Giving PE $2(kQ*Q/√2L)$

Add Them Up to get the Answer


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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