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Two tiny conducting balls of identical mass \(m\) and identical charge \(q\) hang from nonconducting threads of length \(L\). Assume that \(\theta\) is so small that \(\tan \theta\) can be replaced by its approximate equal, \(\sin \theta\). (a) Show that $$ x=\left(\frac{q^{2} L}{2 \pi \varepsilon_{0} m g}\right)^{1 / 3} $$ gives the equilibrium separation \(x\) of the balls. (b) If \(L=120 \mathrm{~cm}, m=10 \mathrm{~g}\), and \(x=5.0 \mathrm{~cm}\), what is \(|q| ?\)

Short Answer

Expert verified
(a) Solved equation confirms given formula. (b) |q| = 4.1 x 10^-8 C.

Step by step solution

01

Analyze the Forces Involved

Each charged ball is subjected to three forces: the tension in the thread (T), the gravitational force (mg) pulling it downward, and the electrical repulsion (F_e) pushing it sideways.The angle of deflection is given by \( \theta \), which is small, so \( \tan \theta \approx \sin \theta \).
02

Derive Trigonometric Relationships

The horizontal component of the tension balances the electric force:\[T \sin \theta = F_e = \frac{k q^2}{x^2}.\]The vertical component balances the gravitational force:\[T \, \cos \theta = mg.\]For small \( \theta \), use \( \sin \theta \approx \tan \theta = \frac{x}{2L} \).
03

Equating Components and Simplifying

From earlier equations:\[T \sin \theta = \frac{kq^2}{x^2} \quad \text{and} \quad T \cos \theta = mg.\]Divide these:\[\tan \theta = \frac{F_e}{mg} \Longrightarrow \frac{x}{2L} = \frac{kq^2}{mgx^2}.\]
04

Solve for Equilibrium Separation (x)

Rearrange the above equation for x:\[x^3 = \frac{kq^2L}{2mg}.\]Substitute \(k = \frac{1}{4 \pi \varepsilon_0}\) to get:\[x^3 = \frac{q^2L}{2 \pi \varepsilon_0 mg},\]which leads to the solution:\[x = \left( \frac{q^2L}{2 \pi \varepsilon_0 mg} \right)^{1/3}.\]
05

Calculate Magnitude of Charge (q)

Given \(x = 0.05 \text{ m}\), \(L = 1.2 \text{ m}\), and \(m = 0.01 \text{ kg}\). Solve:\[q = \sqrt{ \frac{2 \pi \varepsilon_0 mg x^3}{L} }.\]Substitute values:\[q = \sqrt{ \frac{2 \pi (8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2)(0.01 \text{ kg})(9.81 \text{ m/s}^2)(0.05 \text{ m})^3}{1.2 \text{ m}} }\approx 4.1 \times 10^{-8} \text{ C}.\]

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics, describing the force between two point charges. It states that the electrostatic force (\( F_e \) ) is directly proportional to the product of the magnitudes of the charges (\( q_1 \) and \( q_2 \) ) and inversely proportional to the square of the distance (\( r \) ) between them. Mathematically, it is expressed as:\[F_e = k \frac{|q_1 q_2|}{r^2}\]where
  • \( k \) is the Coulomb's constant, equal to \( \frac{1}{4\pi \varepsilon_0} \), with \( \varepsilon_0 \) being the permittivity of free space.
  • The forces are along the line joining the two charges.
In our exercise, the force between the two balls arises due to their identical charges, causing a repulsion that we calculate using the above principle. The force between them can thus be expressed as \( F_e = \frac{k q^2}{x^2} \), where \( x \) is their separation.
Equilibrium in Electrostatic Systems
In an electrostatic system, equilibrium occurs when all forces acting on an object sum to zero, resulting in no net movement. For our setup, each charged ball hangs in equilibrium under the influence of three forces:
  • Gravitational force pulling down, \( mg \).
  • Electrical repulsion force pushing sideways, \( F_e \).
  • Tension in the thread, balancing both the vertical and horizontal components of other forces.
To achieve equilibrium, particularly in a static system, the forces must balance such that the tension's vertical component balances the gravitational force, and its horizontal component matches the electric force:\[T \cos \theta = mg\]\[T \sin \theta = F_e\]In this scenario, understanding these balancing points helps determine the equilibrium separation \( x \) of the balls in the equation \( x = \left( \frac{q^2 L}{2 \pi \varepsilon_0 m g} \right)^{1 / 3} \). By setting these components equal, students can comfortably solve for unknown quantities.
Trigonometric Approximations
Trigonometric approximations are incredibly useful when dealing with small angles, as it simplifies complex equations. When the angle \( \theta \) is small, we assume \( \sin \theta \approx \tan \theta \). In terms of our exercise, this approximation helps us because:- \( \sin \theta \) represents the ratio of the horizontal separation distance to the total length of the string, given by \( \frac{x}{2L} \).- Simplifying \( \tan \theta \) and \( \sin \theta \) allows us to work with simpler linear approximations, which are easier to handle mathematically.This approximation gives us more manageable equations to balance the forces, particularly when substituting values into the force balance equations derived earlier in the equilibrium section. It transforms the complex trigonometric components into relatively straightforward linear relationships.
Electric Force Analysis
Analyzing the electric forces in this context involves considering how these forces interact with other forces present in the system. The electric force analysis starts by recognizing that the electric repulsion between similarly charged objects is what causes them to push apart. The formula for electric force derived from Coulomb's Law in this exercise is:\[F_e = \frac{kq^2}{x^2}\]where \( k \) is a constant and \( x \) is the distance separating the charges. By understanding these interactions, we can rearrange our equations based on the system's equilibrium conditions and solve for unknown variables. For instance, by balancing \( F_e \) with the tension's horizontal component, students derive the relationship between distance \( x \) and charge \( q \):\[\tan \theta = \frac{F_e}{mg} \]Ultimately, this step guides us to determine the unknown charge \( q \) through substitution and simplification, a fundamental aspect of the broader electric force analysis process.

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Most popular questions from this chapter

A charge of \(6.0 \mu \mathrm{C}\) is to be split into two parts that are then separated by \(3.0 \mathrm{~mm}\). What is the maximum possible magnitude of the electrostatic force between those two parts?

We know that the negative charge on the electron and the positive charge on the proton are equal. Suppose, however, that these magnitudes differ from each other by \(0.00010 \%\). With what force would two copper coins, placed \(1.0 \mathrm{~m}\) apart, repel each other? Assume that each coin contains \(3 \times 10^{22}\) copper atoms. (Hint: \(\mathrm{A}\) neutral copper atom contains 29 protons and 29 electrons.) What do you conclude?

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If a cat repeatedly rubs against your cotton slacks on a dry day, the charge transfer between the cat hair and the cotton can leave you with an excess charge of \(-2.00 \mu \mathrm{C}\). (a) How many electrons are transferred between you and the cat? You will gradually discharge via the floor, but if instead of waiting, you immediately reach toward a faucet, a painful spark can suddenly appear as your fingers near the faucet. (b) In that spark, do electrons flow from you to the faucet or vice versa? (c) Just before the spark appears, do you induce positive or negative charge in the faucet? (d) If, instead, the cat reaches a paw toward the faucet, which way do electrons flow in the resulting spark? (e) If you stroke a cat with a bare hand on a dry day, you should take care not to bring your fingers near the cat's nose or you will hurt it with a spark. Considering that cat hair is an insulator, explain how the spark can appear.

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