Find the Potential energy of a system of charges

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Homework Help Overview

The discussion revolves around calculating the potential energy (PE) of a system of charges, specifically focusing on a scenario involving six pairs of charges, three positive and three negative. Participants are examining the implications of their calculations and the correctness of their results.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of potential energy using the formula involving the product of charge pairs and constants. Questions arise regarding the correctness of arriving at a net zero value and the implications of symmetry in the charge distribution.

Discussion Status

The discussion is ongoing, with participants questioning the validity of their calculations and the homework's feedback. Some express uncertainty about the requirement of symmetry in the charge arrangement, while others reflect on the potential errors in their understanding or the homework itself.

Contextual Notes

There are references to homework guidelines that may have been incorrect in previous instances, leading to skepticism about the current assessment of the problem. Participants are also considering the implications of symmetry in their calculations.

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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