Question

a charge Q of mass M is placed at distance D from infinite length of wire of lemda charge density .then its slightly displaced from equilibrium . its motion will be SHM or NOT.

Answer 1

We will denote $x$ a distance between a charge and a wire. Let's study the potential energy of wire:

$V= V_{EM} + V_{gravity}$ , where $V_{EM}$ is energy associated to the electromagnetical interaction between the wire and the charge, $V_{gr}$ is a potential energy associated to gravity. Now to determine the behaviour of a charge due to a small displacement we need to consider the developpement of $V$ in a power series of $x$ until second order :

$V = V(D) + \frac{d}{dx}(V_{gr}+V_{EM})|_{x=D} (x-D) + \frac{1}{2}\frac{d^2}{dx^2}(V_{gr}+V_{EM})|_{x=D} (x-D)^2$

The 0 order term is a constant, therefore it does not affect the behaviour. First order term is zero, as the expression in the brackets is a total force acting on a charge, and charge is in an equilibrium.

The potential energy $V_{EM}$ depends only on $x$ due to the symmetries of an infinite wire (rotation and translation parallel to wire). First derivative of $V_{EM}$ is minus the force, therefore let's calculate the field of an infinite wire. We can apply Gauss theorem to find that $E_{wire} = \frac{\lambda}{2\pi\epsilon_0 x}$ . Therefore $\frac{d^2}{dx^2} V_{EM} =\frac{d}{dx}(-QE_{wire})= \frac{Q\lambda}{2\pi\epsilon_0x^2}$ . Therefore this will be a simple harmonic motion if

$-(\frac{d}{dx}Mg)|_{x=D} + \frac{Q\lambda}{2\pi\epsilon_0 D^2}>0$ . If we suppose that $g$ does not depend on a distance (which is a pretty good approximation), this will be a simple harmonic motion if $Q\cdot\lambda>0$ , i.e. the charge and the wire's charge density are of the same sign (in this case charge is above the wire).

If the displacement is made in a direction that is parallel to wire, charge stays at equilibrium.

If the displacement have a non-zero component in a direction orthogonal to wire axis and distance to a wire, the equilibrium is unstable and motion will now be a SHM.