Respuesta :
Answer:
the magnitude of the force is 192.29 N
Solution:
As per the question:
Charges present on the corner of the triangle are same, Q = [tex]- 3.1\times 10^{- 9} C[/tex]
Since, its an equilateral triangle, distance between the charges, l = 0.50 m
Now,
The Coulomb force on a charge due to the other is:
[tex]F_{C} = K\frac{Q^{2}}{l^{2}}[/tex]
where
K = Coulomb constant = [tex]9\times 10^{9} C^{2}/m^{2}[/tex]
[tex]F_{C} = (9\times 10^{9})\frac{(3.1\times 10^{- 9})^{2}}{0.5^{2}}[/tex]
[tex]F_{C} = 111.6 N[/tex]
The the net force on the charges in an equilateral triangle on all the charges due to each other:
[tex]F_{eq} = \sqrt{3}F_{C} = \sqrt{3}\times 111.6 = 193.29 N[/tex]
