Question #154263

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.


1
Expert's answer
2021-01-11T07:39:29-0500

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

V=VEM+VgravityV= V_{EM} + V_{gravity} , where VEMV_{EM} is energy associated to the electromagnetical interaction between the wire and the charge, VgrV_{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 VV in a power series of xx until second order :

V=V(D)+ddx(Vgr+VEM)x=D(xD)+12d2dx2(Vgr+VEM)x=D(xD)2V = 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 VEMV_{EM} depends only on xx due to the symmetries of an infinite wire (rotation and translation parallel to wire). First derivative of VEMV_{EM} is minus the force, therefore let's calculate the field of an infinite wire. We can apply Gauss theorem to find that Ewire=λ2πϵ0xE_{wire} = \frac{\lambda}{2\pi\epsilon_0 x} . Therefore d2dx2VEM=ddx(QEwire)=Qλ2πϵ0x2\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

(ddxMg)x=D+Qλ2πϵ0D2>0-(\frac{d}{dx}Mg)|_{x=D} + \frac{Q\lambda}{2\pi\epsilon_0 D^2}>0 . If we suppose that gg does not depend on a distance (which is a pretty good approximation), this will be a simple harmonic motion if Qλ>0Q\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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