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Chapter 2: Problem 2

A short insulating rod of negligible mass carries a small charged body at one end. The rod is pivoted about its centre so that it can rotate about a vertical axis. A fixed charge of opposite sign is placed in the same horizontal plane as the rod and at a large distance \(d\) from it. Show that if Coulomb's law holds, small oscillations of the rod would be simple harmonic with a period proportional to \(d\).

Short Answer

Expert verified
Question: Prove that for a short insulating rod with a charged body at one end and a fixed charge of opposite sign at a large distance 'd', small oscillations of the rod would be simple harmonic with a period proportional to 'd', assuming Coulomb's law holds. Answer: The period of oscillation, T, for this system is given by \(T = \frac{2\pi}{\sqrt{\frac{k|q_1q_2|}{md^2}}}\), which shows that the period is proportional to the distance 'd' between the charged objects.

Step by step solution

01

Set up the system and find the net force

Let the charge on the rod be \(q_1\) and the fixed charge be \(q_2\). They are at a distance '\(d\)' apart. According to Coulomb's law, the electrostatic force between these two charges is given by \(F_e = \frac{k|q_1q_2|}{r^2}\), where \(r\) is the distance between the charges, and \(k\) is Coulomb's constant. In our case, \(r \approx d\). Therefore, \(F_e \approx \frac{k|q_1q_2|}{d^2}\)
02

Analyze the force for small oscillations

When the rod undergoes small oscillations, let the displacement angle be \(\theta\). For small angles, we can assume \(\sin\theta \approx \theta\). Since the electrostatic force acts along the line joining the charges, we need to find the component of the force acting perpendicular to the rod to restore it back to its original position. This component is \(F_\perp = F_e\sin\theta \approx F_e\theta\).
03

Derive an equation for the restoring force

The restoring force acting on the charged body in the rod is given by \(F_\perp = m\omega^2\theta\), where \(m\) is the mass of the charged body and \(\omega\) is the angular frequency of oscillation. Comparing this with the previous equation, we get \(F_e\theta \approx m\omega^2\theta\). Canceling the \(\theta\) term from both sides, we have: \(F_e \approx m\omega^2\)
04

Calculate the period of oscillation

The period of oscillation is given by \(T = \frac{2\pi}{\omega}\). From the previous equation, we can solve for \(\omega\): \(\omega^2 = \frac{F_e}{m}\) \(\omega = \sqrt{\frac{F_e}{m}}\) Substituting this into the equation for the period, we get: \(T = \frac{2\pi}{\sqrt{\frac{F_e}{m}}}\) Now substitute the expression for \(F_e\) that we derived in Step 1: \(T = \frac{2\pi}{\sqrt{\frac{k|q_1q_2|}{md^2}}}\) Finally, we see that the period of oscillation, \(T\), is proportional to \(d\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Imagine two charges in space, like stars in the night sky, interacting with an invisible force. Coulomb's law is the scientific equivalent of a recipe for understanding how these charges attract or repel each other. It tells us that the force (\(F_e\)) between two point charges is directly proportional to the product of their charges (\(q_1\text{ and }q_2\text{, respectively}\text{ and inversely proportional to the square of the distance (\(d^2\text{, in the equation above}\text{ between them. Coulomb's constant (\(k\text{ stands as a universal proportionality factor in this formula, allowing us to calculate the electrostatic force with precision.\text{\text{\text{br}\br}\text{ul}\li}\text{Coulomb's law formula: }\text(\text(\text(\text(\text(\text(\text(\text(\text(\text(\}}}}}}}}}}}}}}}}}intValue()).equals(

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Most popular questions from this chapter

A horizontal thin rod of length \(L\) possesses a charge \(\lambda\) per unit length distributed uniformly along it. A point charge \(Q\) is situated vertically above one end of the rod at a height \(h\) above it. What force does \(Q\) exert on the rod?

Two fixed point charges \(-Q\) are a distance \(2 a\) apart and a third point charge \(+Q\) is situated midway between them. Find the potential energy of the system when \(+Q\) is displaced a distance \(x\) along the line of charges, taking the zero for \(U\) when \(+Q\) is at the midpoint. Investigate the stability of \(+Q\) when \(x=0 .\)

Compare the electric and gravitational forces between two electrons or between two protons a certain distance apart. Why do gravitational forces predominate at astronomical distances and electric forces at atomic distances?

Ionic crystals consist of a three-dimensional array of positive and negative ions and the calculation of its potential energy due to Coulomb forces is tedious. As a simpler example, consider a one-dimensional array consisting of a straight row of \(N\) point charges, alternately \(+e\) and \(-e\) and each a distance \(a\) from its nearest neighbours. If \(N\) is very large, find the potential energy of a charge in the middle of the row and of one at the end in the form \(\alpha e^{2} /\left(4 \pi \varepsilon_{0} a\right)\). This sets upper and lower bounds to the Coulomb energy of the whole array. \(\alpha\) is known as the Madelung constant for such systems.

Point charges \(+4 Q\) and \(-Q\) are separated by a distance \(a\). Show that the only positions where a third charge \(+Q\) could be in equilibrium are along the line joining the first two, and find any such positions. Investigate the stability of the equilibrium by finding the direction of the force for a small displacement of \(+Q\)
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