Point charges in a regular hexagon

In summary, the provided answer is (2×sqrt3×k×q^2)/a^2, and the value of 'a' is the hexagon's side-length, while 'R' should be the distance from an outer charge to the centre. Additionally, considering the x and y components of each of the 6 forces on the centre charge and using symmetry can simplify the solution.
  • #1
rbh
9
1
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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  • #2
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:
  • #3
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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