Point charges in a regular hexagon

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SUMMARY

The correct formula for the force between point charges arranged in a regular hexagon is given by (2×sqrt3×k×q^2)/a^2. The variable 'a' represents the side length of the hexagon, while 'R' is the distance from an outer charge to the center. The calculation of 'R' as (sqrt3 × a)/2 is incorrect. Analyzing the x and y components of the forces acting on the center charge, along with symmetry, simplifies the solution significantly.

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  • Understanding of electrostatic forces and Coulomb's law
  • Familiarity with geometric properties of regular hexagons
  • Knowledge of vector components in physics
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Students and professionals in physics, particularly those focusing on electrostatics, as well as educators teaching concepts related to forces and geometry in a hexagonal arrangement.

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Homework Statement
Regular hexagon with side length a, has q,q,q,q,-q,-q point charges in vertices. What force would point charge q expierence if it was put in a hexagon center?
Relevant Equations
F=(kq^2)/R^2
R=(sqrt3 × a)/2
The answer should be (2×sqrt3×k×q^2)/a^2. What did I do wrong?
 

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I can't follow your working but...

Is 'a' the hexagon's side-length? Is 'R' the distance from an outer charge to the centre? If so, 'R=(sqrt3 × a)/2' is wrong!

Also, have you thought about the x and y components of each of the 6 forces on the centre charge? Then, using symmetry, the solution becomes very simple.

Edit - typo' corrected.
 
Last edited:
Steve4Physics said:
I can't follow your working but...

Is 'a' the hexagon's side-length? Is 'R' the distance from an outer charge to the centre? If so, 'R=(sqrt3 × a)/2' is wrong!

Also, have you thought about the x and y components of each of the 6 forces on the centre charge? Then, using symmetry, the solution becomes very simple.

Edit - typo' corrected.
Oh yeah, I mixed it up with inscribed circle radius, thanks.
 

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