Lecture Notes: Force on a Charge at the Centroid of an Equilateral Triangle

Context

• Consider three charges ( q_1, q_2, q_3 ) each equal to ( q ).
• These charges are placed at the vertices of an equilateral triangle ABC with side length ( l ).
• A charge ( Q ) is placed at the centroid ( O ) of the triangle.

Analysis

• Geometry of the Setup:

  • The equilateral triangle has sides of length ( l ).
  • A perpendicular ( AD ) is drawn to the side ( BC ).
  • The length of ( AD ) is calculated as ( AD = AC \cdot \cos(30^\circ) = \frac{\sqrt{3}}{2}l ).
  • The distance from a vertex to the centroid, ( AO ), is ( \frac{2}{3}AD = \frac{l}{3} ).
  • By symmetry, ( AO = BO = CO ).

• Forces on Charge ( Q ):

  • Force ( F_1 ) on ( Q ) due to ( q_1 ) at ( A ):
    • ( F_1 = \frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2} ) directed along ( AO ).
  • Force ( F_2 ) on ( Q ) due to ( q_2 ) at ( B ):
    • ( F_2 = \frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2} ) directed along ( BO ).
  • Force ( F_3 ) on ( Q ) due to ( q_3 ) at ( C ):
    • ( F_3 = \frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2} ) directed along ( CO ).

• Resultant Force:

  • The resultant of ( F_2 ) and ( F_3 ) is along ( OA ), given by the parallelogram law.
  • Total force on ( Q ) is given by:
    • ( F_{total} = \frac{3}{4\pi\varepsilon_0} \frac{Qq}{l^2}(\hat{r} - \hat{r}) = 0 ).
  • Due to symmetry, all three forces sum to zero._

Conclusion

• The forces due to each charge on ( Q ) symmetrically cancel out.
• The net force on the charge ( Q ) at the centroid is zero.

Symmetry Argument

• If the resultant force was non-zero and the system was rotated through 60 degrees about ( O ), the symmetry would still ensure that the net force remains zero.

This setup illustrates the concept of symmetry in electrostatic forces and how it leads to the cancellation of forces in equilateral arrangements.