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Three 5.00-g Styrofoam balls of radius \(2.00 \mathrm{~cm}\) are coated with carbon black to make them conducting and then are tied to 1.00 -m-long threads and suspended freely from a common point. Each ball is given the same charge, q. At equilibrium, the balls form an equilateral triangle with sides of length \(25.0 \mathrm{~cm}\) in the horizontal plane. Determine \(q\)

Short Answer

Expert verified
Answer: The charge on each Styrofoam ball is ±8.23 × 10⁻⁵ C.

Step by step solution

01

Draw a diagram and identify forces

Draw a diagram that includes all three Styrofoam balls, threads, and the forces acting on each ball. There are two forces acting on each charged ball: Tension in the thread and electrostatic repulsion force between the charged balls.
02

Applying Coulomb's Law

The electrostatic force between two charged objects can be found using Coulomb's Law: \(F_e = k \frac{q_1 q_2}{r^2}\), where \(F_e\) is the force between the objects, \(k\) is the electrostatic constant (\(8.99 x 10^9 \mathrm{N m^2/C^2}\)), \(q_1\) and \(q_2\) are the charges of the objects, and \(r\) is the distance between them. In our case, each ball has the same charge (\(q\)), and the distance between them is \(25.0 \mathrm{~cm}\) or \(0.25 \mathrm{~m}\).
03

Applying Newton's Second Law and Trigonometry

Newton's Second Law tells us that the sum of forces acting on an object in equilibrium is zero. For each Styrofoam ball, the vertical components of tension in the thread must balance the weight of the ball, and the horizontal components of tension must balance the electrostatic repulsion force. First, let's calculate the vertical and horizontal components of tension: \(T_{\mathrm{vertical}} = T\cos\theta\) \(T_{\mathrm{horizontal}} = T\sin\theta\) where \(\theta\) is the angle between the thread and the vertical axis. To find \(\theta\), we can use trigonometry. In the equilateral triangle formed by the balls, we can draw a vertical line from a vertex to the midpoint of the opposite side, creating a right triangle with a base half the length of a side and a hypotenuse equal to the length of the thread. The angle between this vertical line and the thread is equal to \(\theta\). We can compute \(\tan{\theta}\): \(\tan\theta = \frac{0.25/2}{\sqrt{1^2 - (0.25/2)^2}}\) Now, we can use the sum of forces to find the tension in the thread: \(T_{\mathrm{vertical}} = T\cos\theta = mg\) \(T_{\mathrm{horizontal}} = T\sin\theta = F_e\)
04

Solving for q

First, we calculate the tension \(T\) from the vertical component equation: \(T = \frac{mg}{\cos\theta}\) Now, substitute the value of T in the equation for the horizontal component: \(F_e = T\sin\theta = \frac{mg\sin\theta}{\cos\theta}\) Rearrange the equation to solve for \(q\): \(q^2 = \frac{mg\sin\theta r^2}{k\cos\theta}\) Substitute the values and solve for \(q\): \(q^2 = \frac{(5 \times 10^{-3}\mathrm{kg})(9.8\mathrm{m/s^2})(\sin\theta)(0.25\mathrm{m})^2}{(8.99 \times 10^9\mathrm{N m^2/C^2})(\cos\theta)}\) \(q = \pm\sqrt{6.78\times10^{-8}C^2}\) \(q = \pm8.23 \times 10^{-5}\mathrm{C}\) So the charge on each Styrofoam ball is \(\pm 8.23 \times 10^{-5}\mathrm{C}\).

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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 an essential principle in electrostatics that describes the force of interaction between two charged objects. It states that the electrostatic force (
  • Is directly proportional to the product of the magnitudes of the two charges (\(q_1\) and \(q_2\)).
  • Is inversely proportional to the square of the distance (\(r\)) between the charges.
The law is mathematically represented as:\[F_e = k \frac{q_1 q_2}{r^2}\]where \(k\) is Coulomb's constant (\(8.99 \times 10^9\) N m²/C²), and \(F_e\) is the electrostatic force. In the provided exercise, all three Styrofoam balls have the same charge (q), making them repel each other equally due to their identical charges. By knowing the distance of 0.25 m between any two balls, Coulomb's Law helps in determining the force of repulsion exerted between each pair of balls. This force plays a crucial part in maintaining the equilibrium.
  • Use Coulomb's Law to calculate the interaction forces in electrostatic systems.
  • Understand the role of charge magnitude and distance in determining the strength of electrostatic forces.
This principle allows us to understand how forces acting on the charged objects work together to form an equilateral triangle, as in the exercise.
Newton's Second Law
Newton's Second Law of Motion expresses the relationship between the forces acting on an object and the motion that results from these forces. It is formally stated as:\[F_{net} = ma\]where \(F_{net}\) is the net force acting on the object, \(m\) is the mass of the object, and \(a\) is the acceleration. In the case of the Styrofoam balls in an equilateral triangle setup, the balls are in equilibrium so their acceleration is zero. This implies:\[T_{horizontal} = F_e\]\[T_{vertical} = mg\]Here, the tension in the thread (\(T\)) is split into horizontal and vertical components. The vertical component balances the gravitational force, while the horizontal component balances the electrostatic force.
  • The sum of forces in equilibrium is zero because the system is not accelerating.
  • Understanding the balance of forces helps us figure out the role of tension in both maintaining equilibrium and counteracting electrostatic and gravitational forces.
Using these ideas, we can find accurate relations between force components, which are vital for solving the given exercise's problem.
Equilateral Triangle
An equilateral triangle is a significant geometric shape where all three sides are equal in length, and all angles are equal to 60 degrees. In the context of the exercise, the three Styrofoam balls form an equilateral triangle in the horizontal plane, with equal sides of 0.25 m.
  • Each side of an equilateral triangle is identical in length, which provides symmetry in the problem.
  • The uniform angles of 60 degrees simplify trigonometric calculations involved in determining forces.
In this setup, the vertical and horizontal components of tension in the strings can be calculated using trigonometry. Trigonometric functions such as sine and cosine rely on these angles to help split the tension (\(T\)) into components that balance gravitational and electrostatic forces.
  • The equal sides ensure that each ball experiences identical forces, contributing to the system's equilibrium.
  • Understanding the properties of an equilateral triangle aids in solving the equilibrium problems encountered in the exercise.
Equilateral geometry not only ensures a symmetric force distribution but also simplifies the equations necessary to find the electrostatic charge (q) in the problem.

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Most popular questions from this chapter

A charge \(Q_{1}\) is positioned on the \(x\) -axis at \(x=a\). Where should a charge \(Q_{2}=-4 Q_{1}\) be placed to produce a net electrostatic force of zero on a third charge, \(Q_{3}=Q_{1}\), located at the origin? a) at the origin c) at \(x=-2 a\) b) at \(x=2 a\) d) at \(x=-a\)

A metal plate is connected by a conductor to a ground through a switch. The switch is initially closed. A charge \(+Q\) is brought close to the plate without touching it, and then the switch is opened. After the switch is opened, the charge \(+Q\) is removed. What is the charge on the plate then? a) The plate is uncharged. b) The plate is positively charged. c) The plate is negatively charged. d) The plate could be either positively or negatively charged, depending on the charge it had before \(+Q\) was brought near.

A small ball with a mass of \(30.0 \mathrm{~g}\) and a charge of \(-0.200 \mu \mathrm{C}\) is suspended from the ceiling by a string. The ball hangs at a distance of \(5.00 \mathrm{~cm}\) above an insulating floor. If a second small ball with a mass of \(50.0 \mathrm{~g}\) and a charge of \(0.400 \mu \mathrm{C}\) is rolled directly beneath the first ball, will the second ball leave the floor? What is the tension in the string when the second ball is directly beneath the first ball?

Two charged objects experience a mutual repulsive force of \(0.10 \mathrm{~N}\). If the charge of one of the objects is reduced by half and the distance separating the objects is doubled, what is the new force?

Two charged spheres are initially a distance \(d\) apart. The magnitude of the force on each sphere is \(F\). They are moved closer to each other such that the magnitude of the force on each of them is \(9 F\). By what factor has the difference between the two spheres changed?

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