what is the explanation for this?

Solution:
Let the charge be q and length of each side of the equilateral triangle be a.
Resultant force on one of the charge due to the other two,

F1=Kq2a2 and F2=Kq2a2 F'A=F12+F22+2F1F2cos60°=3×Kq2a2Simillarly by symmetry,F'B=3×Kq2a2andF'C=3×Kq2a2Now by symmetry, vector sum of all these forces,F'A+F'B+F'C=0



Note: It is important to note that all the forces are acting on different particles, so it can not be added, but the solution is as per the question asked for the sum of these forces.

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