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Question

Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C of mass m and charge q is kept at the middle point of the line AB. (a) If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it? (b) Assuming x<<d, show that this force is proportional to x. (c) Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement? Find the time period of the oscillations if these conditions are satisfied.

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Solution

(a) The charge q is displaced by a distance x on the perpendicular bisector of AB.
As shown in the figure, the horizontal component of the force is balanced.

sinθ=xd22+x2
Total vertical component of the force, F'=2Fsinθ
F'=2×14πε0×qQd22+x2×xd22+x2F'=12πε0×qQxd22+x23/2
This is the net electric force experienced by the charge q.

(b) When x < < d:
F'=12πε0qQxd23 x2<<d22F'=4πε0qQxd3

(c) For the particle to execute simple harmonic motion:
F' = mw2x
4πε0qQxd3=m2πT2xT2=mπ3ε0d3Qq T =mπ3ε0d3Qq1/2

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