Question

Two negative charges \((-q\) and \(-q)\) of equal magnitude 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 \((1+x)^{n}=1+n x,\) for \(x \ll 1,\) to find an expression for the frequency of the resulting oscillation. (Hint: Keep only terms that are linear in \(\delta .\) )

Short answer

Answer: The frequency of the resulting oscillation is given by the formula: $$ f = \frac{1}{2\pi}\sqrt{\frac{kq^2}{md^3}} $$

Step-by-step solution

Determine the force acting on the positive charge from each negative charge

When the positive charge is at coordinate \((0, \delta)\), the distance between it and each negative charge is given by the Pythagorean theorem. The distances are \(r_1 = \sqrt{(-d)^2 + \delta^2}\) and \(r_2 = \sqrt{d^2 + \delta^2}\). The force acting on the positive charge due to each negative charge is given by Coulomb's law: $$ F_1 = \frac{kq(-q)}{r_{1}^2},\ F_2 = \frac{kq(-q)}{r_{2}^2}, $$ where \(k\) is the electrostatic constant.

Determine the angle between the forces and the y-axis

To find the net force in the y-direction, we must take into account the angle between the forces and the y-axis. The angle, \(\theta\), is the same for both forces, which can be found using trigonometry: $$ \cos{\theta} = \frac{\delta}{\sqrt{d^2 + \delta^2}} $$

Calculate the net force in the y-direction

To calculate the net force in the y-direction, we must find the component of each force acting in the y-direction. The y-components are given by: $$ F_{1y} = F_1 \cos{\theta} = \frac{kq(-q)}{r_{1}^2} \cdot \frac{\delta}{\sqrt{d^2 + \delta^2}},\ F_{2y} = F_2 \cos{\theta} = \frac{kq(-q)}{r_{2}^2} \cdot \frac{\delta}{\sqrt{d^2 + \delta^2}} $$ By combining the y-components, we find the net force in the y-direction: $$ F_{net} = F_{1y} + F_{2y} = \frac{kq^2}{d^2 + \delta^2}(D^2 + \delta^2)^{-3/2} \cdot \delta $$

Calculate the frequency using the binomial expansion

First, we need to simplify our expression for the net force by using the binomial expansion: $$ \frac{1}{(d^2 + \delta^2)^{3/2}} = \frac{1}{(d^2(1 + \frac{\delta^2}{d^2})^{3/2}} = \frac{1}{d^3}(1 - \frac{3}{2}\frac{\delta^2}{d^2}) $$ Now, we can find the net force acting on the positive charge with only linear terms in \(\delta\): $$ F_{net} = \frac{kq^2}{d^3}(1 - \frac{3}{2}\frac{\delta^2}{d^2}) \cdot \delta = \frac{kq^2}{d^3}\delta - \frac{3}{2}\frac{kq^2}{d^5}\delta^2 $$ Applying Hookes's law for a harmonic oscillator, we get: $$ F_{net} = -m\omega^2\delta $$ Now, we can solve for the angular frequency \(\omega\): $$ \omega^2 = \frac{kq^2}{md^3} $$ Finally, we can find the frequency of the oscillation: $$ f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{kq^2}{md^3}} $$ So, the frequency of the resulting oscillation is: $$ f = \frac{1}{2\pi}\sqrt{\frac{kq^2}{md^3}} $$

Key concepts

Coulomb's Law

At the heart of electrostatics lies Coulomb's Law, a fundamental principle that explains how charged particles interact with one another. Coulomb's Law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is: \[ F = k \frac{{|q_1 q_2|}}{{r^2}} \]
Here, \( F \) is the magnitude of the electrostatic force, \( k \) is Coulomb's constant (9 \times 10^9 \text{{ Nm}}^2\text{{/C}}^2\text{{), }} \( q_1 \text{{ and }} q_2 \) are the charges, and \( r \) is the distance between the charges. The force is attractive if the charges are of opposite signs and repulsive if they are of the same sign.

Understanding Coulomb’s Law is crucial when analyzing the movement of the charged particle in our exercise, as this law provides the basis for calculating the forces acting on the positive charge due to the two fixed negative charges.

Binomial Expansion

The Binomial Expansion is a mathematical tool that allows us to expand expressions raised to a power in the form of \[ (1 + x)^n \].

For small values of \( x \), where \( x \ll 1 \), the binomial expansion becomes an approximation: \[ (1 + x)^n \approx 1 + nx \].

This is extremely useful in physics when dealing with quantities that are small compared to the reference values, simplifying complex expressions. In our exercise, the binomial expansion is employed to approximate the expression for the distance between charges because \( \delta \) is much smaller than \( d \). This approximation simplifies the calculation for the net force, which in turn is used to determine the oscillation frequency of the positive charge.

Oscillation Frequency

Oscillation frequency, denoted by \( f \), is a measure of how often a periodic motion repeats itself in a unit of time. It's typically expressed in hertz (Hz), where one Hz equals one oscillation per second.

In harmonic oscillators, such as the positive charge in our exercise, the frequency can be determined through the properties of the system. For a simple harmonic oscillator, the frequency is related to the mass of the oscillating object and the force constant, or stiffness, of the restoring force, which strives to return the object to its equilibrium position.

The formula connecting oscillation frequency, harmonic force, and mass is given by: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
Here, \( k \) is the force constant, and \( m \) is the mass of the oscillator. Applying this in our context, the frequency of the oscillation is determined by solving for the angular frequency \( \omega \) from the net force using Hooke's law and then expressing \( f \) in terms of \( \omega \).