Find the Potential energy of a system of charges

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SUMMARY

The discussion centers on calculating the potential energy (PE) of a system of charges, specifically involving six pairs of charges where three are positive and three are negative. The initial calculation using the formula $$\frac{k}{r}(3q^2 - 3q^2)$$ resulted in zero, which was questioned as incorrect. Participants confirmed that the calculation method was accurate, highlighting the importance of recognizing symmetry in charge distributions, particularly the 3:1 ratio of positive to negative charges. The conversation also noted that the homework prompt may have been misleading, as it has been incorrect in the past.

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  • Understanding of electrostatics and potential energy calculations
  • Familiarity with Coulomb's law and the constant k
  • Knowledge of charge interactions and symmetry in charge distributions
  • Basic algebra for manipulating equations involving charges
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Jaccobtw
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Homework Statement
3 electrons and a proton are arranged in a regular tetrahedron with a side length of 10nm. What is the total electrostatic potential energy of this configuration? Assume the energy is zero when the electrons are infinitely far from each other. Give your answer in J
Relevant Equations
$$k \frac{q_1 q_2}{r}$$
Screenshot (104).png

There are six pairs. three turn out to be negative and three turn out to be positive (3q^2 - 3q^2) which nets zero when you add them together with the equation. But zero was the incorrect answer. Did I do something wrong? Thank you
 
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Jaccobtw said:
But zero was the incorrect answer. Did I do something wrong?
You must have done. How did you try to calculate the PE?
 
PeroK said:
You must have done. How did you try to calculate the PE?
$$\frac{k}{r}(3q^2 - 3q^2)$$ which is zero.
 
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Jaccobtw said:
$$\frac{k}{r}(3q^2 - 3q^2)$$ which is zero.
What is that?!
 
PeroK said:
What is that?!
What's wrong? That's how you calculate the potential energy of a system of charges. You have 6 pairs. 3 of them are q^2 and 3 of them are -(q^2). You multiply them by k and divide by r and add them together to get an answer.
 
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I can't find anything incorrect with your calculation. How do you know that "zero" is incorrect?
 
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Jaccobtw said:
What's wrong? That's how you calculate the potential energy of a system of charges. You have 6 pairs. 3 of them are q^2 and 3 of them are -(q^2). You multiply them by k and divide by r and add them together to get an answer.
Yes, sorry, I failed to see the remarkable symmetry with the 3:1 ratio.
 
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PeroK said:
Yes, sorry, I failed to see the remarkable symmetry with the 3:1 ratio.
I didn't know symmetry was required.
 
kuruman said:
I can't find anything incorrect with your calculation. How do you know that "zero" is incorrect?
The homework says it was wrong, but oddly enough the homework has been wrong many times this year
 
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PeroK said:
Yes, sorry, I failed to see the remarkable symmetry with the 3:1 ratio.
A fun exercise: find the solutions of m protons and n electrons with zero EPE when equidistant (in a space of at least m+n-1 dimensions, of course).
 

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