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Question

Two electrons having charge (e) each are fixed at a distance ‘2d’. A third charge proton placed at the midpoint is displaced slightly by a distance x (x<<d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:
(m= mass of charged particle)

A
(e22πεomd3)1/2
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B
(πεomd32e2)1/2
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C
(2πεomd3e2)1/2
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D
(2e2πεomd3)1/2
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Solution

The correct option is A (e22πεomd3)1/2

Restoring force on proton:-

Fr=2F1sinθ, Where, F1=ke2d2+x2

Fr=2ke2x(d2+x2)3/2

x<<d

Fr=2ke2xd3=Kx

K=2ke2d3=e22πεod3

Angular Frequency :-

ω=Km

ω=e22πεomd3

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