Two equal negative charge -q are fixed at the | vedclass

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(d) By symmetry of problem the components of force on $Q$ due to charges at $A$ and $B$ along $y$-axis will cancel each other while along $x$-axis wil

Vedclass · Accepted Answer

Execute oscillatory but not simple harmonic motion

Vedclass · Answer

(d) By symmetry of problem the components of force on $Q$ due to charges at $A$ and $B$ along $y$-axis will cancel each other while along $x$-axis will add up and will be along $C_O$. Under the action of this force charge $Q$ will move towards $O$. If at any time charge $Q$ is at a distance $x$ from $O$. Net force on charge $Q$

${F_{net}} \Rightarrow 2F\cos 	heta $ $ = 2\frac{1}{{4\pi {\varepsilon _0}}}\frac{{ - qQ}}{{({a^2} + {x^2})}}$ $ 	imes \frac{{ x}}{{({a^2} + {x^2})^{1/2}}}$

i.e., ${F_{net}} =  - \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{2qQx}}{{{{({a^2} + {x^2})}^{3/2}}}}$

As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude $2a$) but not simple harmonic.

Two equal negative charge $-q$ are fixed at the fixed points $(0,\,a)$ and $(0,\, - a)$ on the $Y$-axis. A positive charge $Q$ is released from rest at the point $(2a,\,0)$ on the $X$-axis. The charge $Q$ will

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