Question

Two equal magnitude negative charges \((-q\) and \(-q)\) are fixed at coordinates \((-d, 0)\) and \((d, 0) .\) A positive charge of the same magnitude, \(q\), and with mass \(m\) is placed at coordinate \((0,0),\) midway between the two negative charges. If the positive charge is moved a distance \(\delta \ll d\) in the positive \(y\) -direction and then released, the resulting motion will be that of a harmonic oscillator-the positive charge will oscillate between coordinates \((0, \delta)\) and \((0,-\delta)\). Find the net force acting on the positive charge when it moves to \((0, \delta)\) and use the binomial expansion \((1+x)^{n} \approx 1+n x,\) for \(x \ll 1,\) to find an expression for the frequency of the resulting oscillation.

Short answer

Answer: The frequency of oscillation can be found using the expression \(\omega \approx \sqrt{\frac{2kq^2}{m(d^2+\delta^2)}(1-\frac{1}{2}(\frac{\delta}{d})^2)}\), where \(k\) is the electrostatic constant, \(q\) is the charge of the particles, \(m\) is the mass of the positive charge, \(d\) is the distance between the negative charges, and \(\delta\) is the displacement of the positive charge.

Step-by-step solution

1. Calculate the force acting on the positive charge due to the negative charges

When the positive charge is at the coordinate \((0, \delta)\), it will experience a force from each of the fixed negative charges. Using Coulomb's law, we can find the force acting on the positive charge due to each of the negative charges: \(F_{1}=\frac{k(-q)q}{[(0+d)^2+\delta^2]}=\frac{-kq^2}{(d^2+\delta^2)}\) \(F_{2}=\frac{k(-q)q}{[(0-d)^2+\delta^2]}=\frac{-kq^2}{(d^2+\delta^2)}\) Here, \(F_{1}\) and \(F_{2}\) are the forces acting on the positive charge due to the negative charges at coordinates \((-d, 0)\) and \((d, 0)\) respectively, and \(k\) is the electrostatic constant. Notice that the magnitudes of the forces are the same, but they have different directions.

2. Find the net force acting on the positive charge

The net force acting on the positive charge will be the vector sum of the forces \(F_{1}\) and \(F_{2}\). Let's choose x and y axes such that the negative charges are on the x-axis and the positive charge moves along the y-axis. The horizontal components of the forces cancel each other out since the magnitudes of the forces are the same and their directions are symmetrical about the y-axis. Therefore, only the vertical components of the forces need to be considered. The angle \(\theta\) between the vertical axis and the forces can be calculated as follows: \(\theta = \tan^{-1}(\frac{\delta}{d})\) The vertical component of each force can now be calculated as follows: \(F_{1y} = F_{1} \cos(\theta) = \frac{-kq^2}{(d^2+\delta^2)}\cos(\tan^{-1}(\frac{\delta}{d}))\) As \(\frac{\delta}{d} \ll 1\), we can approximate \(\cos(\tan^{-1}(\frac{\delta}{d}))\) with the binomial expansion: \(\cos(\tan^{-1}(\frac{\delta}{d})) \approx 1 - \frac{1}{2}(\frac{\delta}{d})^2\) Now using this approximate value, we can find \(F_{1y}\): \(F_{1y} \approx -kq^2 (1-\frac{1}{2}(\frac{\delta}{d})^2) \frac{1}{d^2+\delta^2}\) Since the positive charge is symmetrically located between the negative charges, the vertical component of force \(F_{2}\) acting on it will be the same as \(F_{1y}\). The net vertical force, \(F_{net}\), will be the sum of the vertical components of \(F_{1}\) and \(F_{2}\): \(F_{net} = 2 F_{1y} \approx -\frac{2kq^2}{d^2+\delta^2}(1-\frac{1}{2}(\frac{\delta}{d})^2)\)

3. Relate the net force to the motion of a harmonic oscillator

The resulting motion of the positive charge is that of a harmonic oscillator. In this case, the net force acting on the charge is proportional to its displacement from the equilibrium position: \(F_{net} = -k_{eff}\delta\) where \(k_{eff}\) is the effective spring constant for the motion. Now using the expression for the net force found in step 2, we can find \(k_{eff}\): \(k_{eff} \approx \frac{2kq^2}{d^2+\delta^2}(1-\frac{1}{2}(\frac{\delta}{d})^2)\)

4. Find the frequency of oscillation

In a harmonic oscillator, the frequency of oscillation can be related to the effective spring constant and the mass of the oscillating object: \(\omega = \sqrt{\frac{k_{eff}}{m}}\) We can substitute the expression for \(k_{eff}\) from step 3 into this equation to find the frequency of the resulting oscillation \(\omega\): \(\omega \approx \sqrt{\frac{2kq^2}{m(d^2+\delta^2)}(1-\frac{1}{2}(\frac{\delta}{d})^2)}\) This is the expression for the frequency of the resulting oscillation.

Key concepts

Coulomb's Law

Coulomb's Law is a fundamental principle that describes how charges interact with one another through electrostatic force. It states that the force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
\[F = k \frac{|q_1 q_2|}{r^2}\]
Where,

• \(F\) is the magnitude of the electrostatic force between the charges.
• \(q_1\) and \(q_2\) are the magnitudes of the charges.
• \(r\) is the distance separating the charges.
• \(k\) is Coulomb's constant, approximately equal to \(8.99 \times 10^9 \text{Nm}^2/\text{C}^2\).

In scenarios involving multiple charges, like the one in the original exercise, the principle of superposition applies. This means each pair of charges is considered separately, forming individual force vectors that are then combined to calculate the net force acting on a particular charge. The net electrostatic force is determined by adding vectorially all the individual forces from each charge present in the system.

Electrostatic Force

Electrostatic force is the phenomenon that takes place when charged particles exert forces on each other. As stated by Coulomb's Law, these forces can be attractive if the charges are opposite or repulsive if they have like charges. The key factors influencing the electrostatic force are the magnitudes of the charges and the distance between them. In the context of this problem, the positive charge experiences forces from two equal but opposite negative charges located symmetrically about it.

These forces act along the line joining the positive and negative charges. The vector nature of these forces means they have both magnitude and direction, leading to the cancellation of horizontal components and retention of vertical components of the forces acting perpendicularly to the line of charge placement.

For a situation involving multiple forces, the overall or net electrostatic force is the vector sum of the individual forces exerted on the particle of interest. In the exercise, the vertical force components due to symmetry and geometry are crucial in determining the resulting motion and behavioral characteristics of the charge when slightly disturbed.

Binomial Expansion

The binomial expansion is an approximation method that simplifies expressions containing terms raised to a power. It's particularly useful when dealing with small perturbations or changes, where one variable is much smaller than the other. In the context of the provided exercise, binomial expansion simplifies the terms resulting from small deviations of position, particularly when evaluating trigonometric functions or polynomial expressions in denominators.

The binomial theorem states:

• For any integer \(n\) and real numbers \(x\) where \(x \ll 1\), \((1+x)^n \approx 1 + nx\).

This approximation allows us to linearize terms and approximate functions providing an easier path to solution. For example, simplifying \(\cos(\tan^{-1}(\frac{\delta}{d}))\) in the force approximation step using binomial expansion helps achieve a workable expression for the net force, making it easier to derive constants such as the effective spring constant \(k_{eff}\). This, in turn, simplifies determining the frequency of oscillation. Binomial expansion is therefore a powerful tool in physics, especially when dealing with small oscillations and perturbations around equilibrium positions.