The electrostatic force is the electric force that exists between two static and charged particles. Depending on whether the particles are positively or negatively charged, the nature of this force might be attractive or repulsive. Coulomb’s law gives us the equation that connects the force and the charged particles, as well as the distance between them. The length of the equilateral triangle is not stated in this question. As a result, we’ll presume it’s l.
Solution
Now, according to Coulomb’s law,
F=k q1q2/d2
Now, let us calculate the force on first particle 1 say F1 with charge q
We are already aware of that particles with the same charge repel each other and that with opposite charges attract each other. Therefore, F1 is the resultant of repulsive force on particle 1 by 2 say force F12 and attractive force by 3 on 1 say force F13
Now for, the charge on both particles is equal in magnitude and direction and distance is taken as l
Therefore, Coulomb’s law can be modified as
F12 = k q1q2/d2
After substituting the values We get, F12 =k q2/l2
Let, k q2/l2=F …………………………….. (1)
Therefore, |F12 |=F ……………….. (2)
Similarly, for force F13
F13 = k q1q2/d2
After substituting values, We get, F13 =(−k q2/l2) (since charge on particle 3 is negative)
|F13 |=F ……………… (3)
Similarly, if we calculate the magnitude of other forces say F21,F23,F31,F32
We will get the same result
Therefore, |F21|=|F23|=|F31|=|F32|=F ………………. (4)
Now we know that F1 is the resultant of F12 and F13 .
Also, the angle between the two forces is 120 degrees.
Therefore, from (2) and (3) We get, F1 = √F2 + F2 + 2 × F × Fcos120
On solving We get, F1=F …………… (since cos120=−1/2 ) ……………….. (A)
Now, since the distance between the charge 1 and 2 is the same, also their charge and other factors such as angle and magnitude of component forces is the same.
we can say that F1 = F2
Therefore, F2=F ……………………………………. (B)
Now, for charge 3, we can say that the angle between the component vector force say F31 and F32 of force F3 is 60 degrees .
Therefore, the magnitude resultant force F3
after substituting, values of F31 and F32 will be given by,
F3 = √F2 + F2 + 2 × F × Fcos60
On solving, We get, F3 = √3F ………………… (since cos60 =1/2 ) …………… (C)
Now, from (1) we know that kq2l2=F Therefore, from (1), (A) and (B)
We get, F2 = F2 = kq2/l2 Also from (1) and (C We get, F3=√3k q2/l2
Answer
Therefore, force on charged particle q, q and –q is kq2/l2 ,k q2/l2, √3k q2/l2respectively.