The Force acting on charge A will be 0.036N.
Coulomb's law is the force of attraction or repulsion acting on straight line is directly proportional to the charge place on that line and inversly proportional to the square of seperation between them.
F= kq1q2/r²
where F is the electrostatic force ,q are the point charges , K is coulomb constant , r is seperation between the charges.
Value of Coulomb's constant is 8.98×10⁹Nm²/C².
In the question ,we have given the sides of the triangle ,we can consider them as seperation also ,
r=5cm = 5×10-²m
the charge is 10-⁷C.
F= 9×10⁹×1×10-¹⁴/25×10-⁴
F= 0.036N
So, the force acting on Charge A is
0.036 N.
to learn more about Coulomb's Law click herehttps://brainly.com/question/506926
#SPJ9
