Two charges \(-q\) each are fixed separated by distance \(2d.\) A third charge \(q\) of mass \(m\) placed at the mid-point is displaced slightly by \(x(x<<d)\) perpendicular to the line joining the two fixed charged as shown in figure. The time period of the simple harmonic oscillation performed by \(q\) will be:
1. \(T=\left[\dfrac{2\pi^3 \epsilon_0 md^3}{q^2}\right]^{1/2}\)
2. \(T=\left[\dfrac{4\pi^3 \epsilon_0 md^3}{q^2}\right]^{1/2}\)
3. \(T=\left[\dfrac{8\pi^3 \epsilon_0 md^3}{q^2}\right]^{1/2}\)
4. \(T=\left[\dfrac{\pi^3 \epsilon_0 md^3}{2q^2}\right]^{1/2}\)
Let us elaborate the figure fist.
Given, two charge -q at A and B
x = small distance perpendicular to O.
i.e., x<d mass of charge Q is. So, force of attraction at P towards A and B are each where AP=BP=r
Horizontal components of these forces Fn are cancelled out. Vertical components along PO add.
If the net force on q along PO is
When,
where,
i.e., force on charge q is proportional to its displacement from the centre O and it is directed towards O.
Hence, motion of charge q would be simple harmonic, where
and
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