Point charges in a regular hexagon

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The correct formula for the interaction of point charges in a regular hexagon is (2×sqrt3×k×q^2)/a^2. There is confusion regarding the definitions of 'a' as the hexagon's side length and 'R' as the distance from an outer charge to the center. The calculation for 'R' as (sqrt3 × a)/2 is incorrect. Considering the x and y components of the forces on the center charge simplifies the problem using symmetry. Clarification on these points is essential for accurate problem-solving.
rbh
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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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