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Two equal negative charges $-q$ are fixed at points $(0, -a)$ and $(0, a)$ in the $x-y$ plane. A positive charge $Q$ is released from rest at a point $(2a, 0)$. The charge $Q$ will
move to the origin and remain at rest there
move to infinity
execute $SHM$ about the origin
execute oscillatory not $SHM$
Solution
The charge moves towards $x$ – axis. Let at certain instant its position from point $\mathrm{O}$ is $\mathrm{x}$.
Instantaneous resultant force acting over it is given as
${\mathrm{F}_{\mathrm{R}}=2 \mathrm{F} \cos \theta} $
${=2 \frac{\mathrm{qQ}}{4 \pi \varepsilon_{0}\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)} \frac{\mathrm{x}}{\sqrt{\mathrm{a}^{2}+\mathrm{x}^{2}}}} $
${\mathrm{F}_{\mathrm{R}}=\frac{2 \mathrm{q} \mathrm{Q}}{4 \pi \varepsilon_{0}} \frac{\mathrm{x}}{\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}}$
This force is not of the form $\mathrm{F}=\mathrm{kx}$
Therefore, the motion of the charge $\mathrm{Q}$ is oscillatory but not simple harmonic.
Similar Questions
In steady state heat conduction, the equations that determine the heat current $j ( r )$ [heat flowing per unit time per unit area] and temperature $T( r )$ in space are exactly the same as those governing the electric field $E ( r )$ and electrostatic potential $V( r )$ with the equivalence given in the table below.
| Heat flow | Electrostatics |
| $T( r )$ | $V( r )$ |
| $j ( r )$ | $E ( r )$ |
We exploit this equivalence to predict the rate $Q$ of total heat flowing by conduction from the surfaces of spheres of varying radii, all maintained at the same temperature. If $\dot{Q} \propto R^{n}$, where $R$ is the radius, then the value of $n$ is
