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.
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.
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