Consider the charges q, q and –q placed at the vertices of the equilateral triangle, as shown in the figure. What is the force on each charge? - Padhle.Online
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# Consider the charges q, q and –q placed at the vertices of the equilateral triangle, as shown in the figure. What is the force on each charge?

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 = F= kq2/l2 Also from (1) and (C We get, F3=√3k q2/l2

Therefore, force on charged particle q, q and –q is kq2/l2 ,k q2/l2, √3k q2/l2respectively.

Principal, Babu Daudayal SVM, Mathura