Need help - about Coulomb's Law and SHM

[elvisphy]

need help -- about Coulomb's Law and SHM

Two positive charges +Q are held fixed a distance d apart.
A particle of negative chage -q and mass m is placed midway between them, then is given a small displacement perpendicular to the line joining them and released. Show that the particle describes simple harmonic motion and find the period.
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i've found that the net force acting on the -q is
F = - (qQd) / [4(pai)(z)(y^2 + (0.5d)^2)]^(3/2) = ma
where z is the permittivity, and y is the displacement of -q

how i can show that a = - w^2 y?

 

[dextercioby]

Well, you need to redo the calculation of the total net force.

Daniel.

 

[kyle8921]

Hello,

Thank you for reaching out for help with Coulomb's Law and Simple Harmonic Motion (SHM). I am happy to assist you in understanding these concepts.

First, let's review Coulomb's Law. It states that the force between two charged particles is directly proportional to the product of their charges (q and Q) and inversely proportional to the square of the distance between them (d). This is represented by the equation F = kqQ/d^2, where k is the proportionality constant.

Now, let's look at the situation described in your question. We have two fixed positive charges, +Q, held d distance apart, and a negative charge, -q, placed midway between them. When the -q charge is given a small displacement, it will experience a force due to the electric field created by the two fixed charges, according to Coulomb's Law.

Using the equation for Coulomb's Law, we can calculate the net force on the -q charge:

F = kqQ/d^2

= (1/4πε0)(qQ/d^2) (where ε0 is the permittivity of free space)

= (qQd)/(4πε0d^3)

= (qQd)/(4πε0)(d^2)^3

= (qQd)/(4πε0)(y^2 + (0.5d)^2)^3 (since d = 2y)

= ma (according to Newton's Second Law, F = ma)

Therefore, we can see that the acceleration (a) of the -q charge is directly proportional to its displacement (y) and is in the opposite direction. This is the defining characteristic of SHM.

To show that a = -ω^2y, we can rearrange the equation for acceleration:

a = (qQd)/(4πε0)(y^2 + (0.5d)^2)^3

= (qQd)/(4πε0)(y^2 + (0.5d)^2)^2 * (y^2 + (0.5d)^2)^1

= (qQd)/(4πε0)(y^2 + (0.5d)^2)^2 * (y^2 + (0.5d)^2)^1/2

= (qQd)/(4πε0)(