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Two similar negative charges –q are situated at point (0, a) & (0, –a) along Y-axis. A positive charge Q is left from point (2a, 0). Analyse the motion of Q.

1 Answer

Due to symmetry the y components of forces acting on Q due to charges –q at A and B will balance each other and the x components will add up along direction O.
If at any instant Q is at a distance x from O
$F= F_1 \cos q +F_2 \cos q $
$\quad= 2F_1 \cos q [Q F_1 =F_2]$
$\quad= 2 \large\frac{1}{4 \pi \in _0}$$ \bigg[ \large\frac{-qQ}{(a^2+x^2)} \bigg] $$\bigg[ \frac{x}{(a^2+x^2)^{1/2}}\bigg]$
$F= -\large\frac{1}{4 \pi \in_0} \frac{2qQx}{(a^2+x^2)^{3/2}}$ ie $F \mu-x$
Q F $\mu –x,$ the motion is oscillatory having amplitude $2a$ but it will not be S.H.M.
If $ x << a \quad F= -\large\frac{1}{4 \pi \in_0} \frac{2qQ}{a^3} $$x =-kx$
and the motion is S.H.M. with time period T, where
$T= 2 \pi \sqrt { \large\frac{m}{k}}$
$\quad= 2 \pi \sqrt { \large\frac{4 \pi \in_0 ma^3}{2Qq}}$
Hence A is the correct answer.
answered May 29, 2014 by meena.p

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