Problem 67 - University Physics with Modern Physics

Question

CP Two positive point charges \(Q\) are held fixed on the \(x\) -axis at \(x=a\) and \(x=-a .\) A third positive point charge \(q,\) with mass \(m,\) is placed on the \(x\) -axis away from the origin at a coordinate \(x\) such that \(|x| << a\) . The charge \(q,\) which is free to move along the \(x\) -axis, is then released. (a) Find the frequency of oscillation of the charge \(q .\) (Hint: Review the definition of simple harmonic motion in Section \(14.2 .\) Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots,\) valid for the case \(|z|<1 .\) (b) Suppose instead that the charge \(q\) were placed on the \(y\) -axis at a coordinate \(y\) such that \(|y| < < a\) , and then released. If this charge is free to move anywhere in the \(x y\) -plane, what will happen to it? Explain your answer.

Step-by-Step Solution

Answer

(a) The frequency of oscillation is \(f = \frac{1}{2\pi}\sqrt{\frac{4k_e Qq}{a^3 m}}\). (b) The charge will oscillate vertically along the y-axis due to the net restoring force.

Step 1: Understand the Problem Setup

We have two fixed charges, each with charge \(Q\), located on the \(x\)-axis at positions \(x = a\) and \(x = -a\). Another charge \(q\) (mass \(m\)) is placed at a position \(x\) on the \(x\)-axis and has small oscillations. We will first solve for the frequency of the oscillation when \(|x| << a\).

Step 2: Write the Expression for Electric Force

The electric force between point charges is given by Coulomb's law: \(F = k_e \frac{q_1 q_2}{r^2}\), where \(k_e\) is Coulomb's constant.For charge \(q\) at position \(x\), the forces due to charges at \(a\) and \(-a\) are: \[F_{1} = k_e \frac{Qq}{(a-x)^2}\] and \[F_{2} = k_e \frac{Qq}{(a+x)^2}\].

Step 3: Approximate Using Binomial Expansion

Since \(|x| << a\), we can use the binomial expansion:\[(1 \pm z)^{-2} \approx 1 \mp 2z \]Apply this to the denominators:\[(a-x)^{-2} \approx a^{-2}(1 + 2\frac{x}{a})\]\[(a+x)^{-2} \approx a^{-2}(1 - 2\frac{x}{a})\]

Step 4: Compute Net Force and Simplify

The net force on \(q\) is:\[F_{net} = F_1 - F_2 = k_e Qa^{-2}[1 + 2\frac{x}{a}]q - k_e Qa^{-2}[1 - 2\frac{x}{a}]q\]Simplify to get: \[F_{net} = 4k_e \frac{Qq}{a^3}x\].This is of the form \(F = -kx\), which indicates simple harmonic motion (SHM).

Step 5: Find the Frequency from SHM Characteristics

For SHM, the restoring force is \(F = -kx\) with \(k = 4k_e \frac{Qq}{a^3}\). The angular frequency \(\omega\) is given by:\[\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{4k_e Qq}{a^3 m}}\].Thus, the frequency \(f\) is:\[f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{4k_e Qq}{a^3 m}}\].

Step 6: Address the Condition when Charge is on y-axis

When the charge \(q\) is placed on the \(y\)-axis at \(y\), its distances from the charges at \(a\) and \(-a\) are equal. The horizontal components of force due to each fixed charge are equal and opposite, so they cancel out leaving only vertical force components. These lead to net vertical forces, which when combined, result in restoring motion towards \(x=0\), hence leading to oscillations along the \(y\)-axis.

Key Concepts

Coulomb's law

Coulomb's law is a fundamental principle that describes the force between two electric charges. It states that the electric force (F) 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. The mathematical form of Coulomb's law is given as: - \( F = k_e \frac{q_1 q_2}{r^2} \) Here, \( k_e \) represents Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges. Coulomb's law is critical for understanding interactions between charged particles. It's similar to Newton's law of gravitation but applies to electrical forces.

Electric force equations

To analyze how electric charges interact, we use electric force equations based on Coulomb's law. For a scenario with charges fixed on the x-axis and a third charge free to oscillate, the initial forces can be determined using: - \( F_1 = k_e \frac{Qq}{(a-x)^2} \) - \( F_2 = k_e \frac{Qq}{(a+x)^2} \) These equations calculate the electric force exerted by charges positioned at \( x=a \) and \( x=-a \) on a charge \( q \) at position \( x \). The formula involves substituting the respective distances into Coulomb's law to determine the individual forces.

Binomial expansion

The binomial expansion is a mathematical tool that simplifies expressions involving powers. It is especially useful when the expressions have small values in the denominators, allowing easier calculations. In this scenario, since we have \(|x| << a\), small terms in expressions can be expanded using: - \((1 \, \pm \, z)^{-2} \approx 1 \, \mp \, 2z\) For the electric forces given earlier: - \((a-x)^{-2} \approx a^{-2}(1 \, + \, 2\frac{x}{a})\) - \((a+x)^{-2} \approx a^{-2}(1 \, - \, 2\frac{x}{a})\) The binomial expansion simplifies the complexity of expressions, making it easier to derive equations necessary for calculating oscillations.

Oscillation frequency

The oscillation frequency of an object describes how fast it moves back and forth in its repetitive movement, a concept tied to simple harmonic motion (SHM). Here, the charge \( q \) oscillates due to the electric forces between it and the fixed charges. We derive the oscillation frequency from the force equation: - The net force simplifies to: \( F_{net} = 4k_e \frac{Qq}{a^3}x \) - For SHM, the force being proportional to displacement \((-kx)\) indicates that the motion follows a periodic path. The effective spring constant \( k \) here involves electrical terms. the angular frequency \( \omega \) of the charge is: - \( \omega = \sqrt{\frac{4k_e Qq}{a^3 m}} \) The actual frequency \( f \) is given by dividing this by \( 2\pi \): - \( f = \frac{1}{2\pi}\sqrt{\frac{4k_e Qq}{a^3 m}} \) This describes how fast \( q \) oscillates about its equilibrium position.