Electric Charge and Coulomb's Law

[Fusilli_Jerry89]

Coulomb's Law Problem - Please Help

Homework Statement

Two positive charges +Q are held fixed a distance d apart. A particle of negative charge -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 of period sqrt((epsilon not)m((pi)^3)(d^3))/(qQ).

Homework Equations

The Attempt at a Solution

SO I calculated the resultant force on the Q charge at any point and found it to be [((sqrt(2))qQ)]/[4pi(epsilon not)(r^2)). I then saw that F=-kz and T = 2pi*sqrt(m/k).
After plugging everything in and seeing that z was neglible when compared to d/2, I came up with several different answers all the same as sqrt((epsilon not)m((pi)^3)(d^3))/(qQ) only I had coefficients in the numerator and denominator. I have no idea what I am doing wrong? Also, how do you prove it is simple harmonic motion?

Here's what I did:

I said that r is approx. equal to d/2. and z is rsin(theta). After I pug this all into T = 2p*sqrt(m/k) I get T = sqrt([16(pi^3)(epsilon not)m(r^3)sin(theta)]/[sqrt(2)qQ]).
I don't get what I am doing wrong. Plz help...thx.

 

[Shooting Star]

Look at this problem in this forum -- identical.

https://www.physicsforums.com/showthread.php?t=205537

 

[Moetasim]

Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this problem, we are given two fixed positive charges +Q and a negative charge -q placed in between them. We are asked to show that the negative charge will undergo simple harmonic motion with a period of sqrt((epsilon not)m((pi)^3)(d^3))/(qQ).

To solve this problem, we first need to find the resultant force on the negative charge at any point. Using Coulomb's Law, we can calculate the force between the negative charge and each of the positive charges. Since the negative charge is equidistant from both positive charges, the forces will cancel out in the horizontal direction and only the vertical components will contribute to the net force. This means that the resultant force will be equal to the sum of the vertical components of the forces from the two positive charges.

Next, we can use Newton's Second Law (F=ma) to relate the resultant force to the acceleration of the negative charge. Since the negative charge is given a small displacement perpendicular to the line joining the two positive charges, we can assume that its acceleration will be directly proportional to its displacement and in the opposite direction. This is the definition of simple harmonic motion.

Now, we can equate the resultant force to the force in the simple harmonic motion equation (F=-kz) and solve for the spring constant (k). We can then plug in the value of k into the period equation (T=2pi*sqrt(m/k)) and simplify to get the desired result.

Remember to always check your units and make sure they are consistent throughout your calculations. Also, in this problem, we are dealing with a small displacement, so we can use the small angle approximation sin(theta)~theta. This will simplify the final expression and give us the correct answer.

In summary, to show that the negative charge undergoes simple harmonic motion, we need to show that the resultant force on the negative charge can be related to the force in the simple harmonic motion equation and solve for the period of oscillation. By using Coulomb's Law and Newton's Second Law, we can show that the negative charge will indeed undergo simple harmonic motion with a period of sqrt((epsilon not)m((pi)^3)(d^3))/(qQ).