three charges equal to -Q are placed at the three corners of equilateral triangle of side a and charge q at its centre. If the system is at equilibrium then the value of q is?
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Answer:
q = Q/√3
Explanation:
net force on "-Q" at one corner:
(i) due to charges at other corners (away from the centre of the ∆)
F = 2K(Q^2/a^2)cos(30°)
=> F = √3 K(Q^2/a^2)......(1)
(ii) due to charge at the centre (towards the centre)
F' = KqQ/r^2 ,. r = (2/3)(√3/2)a = a/√3
=> F' = 3KqQ/a^2 ..........(2)
since, net force should be zero.
hence, F = F'
=> √3 K(Q^2/a^2) = 3KqQ/a^2
=> q = Q/√3
hope, it helps you.
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