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Two charged particles having charge \(2.0 \times 10^{-8} \mathrm{C}\) each are joined by an insulating string of length \(1 \mathrm{~m}\) and the system is kept on a smooth horizontal table. Find the tension in the string.

Short Answer

Expert verified
The tension in the string is \(3.596 \times 10^{-6} \text{ N}\).

Step by step solution

01

Understand the Problem

We have two charged particles with charges \(q = 2.0 \times 10^{-8} \text{ C}\) each, connected by a string of length \(1 \text{ m}\). The particles are on a smooth, horizontal table. We need to find the tension in the string, which will be due to the electrostatic force of repulsion between the charges.
02

Recall Coulomb's Law

The electrostatic force \(F\) between two point charges is given by Coulomb's Law: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where:- \(k\) is the Coulomb's constant, approximately \(8.99 \times 10^9 \text{ N m}^2/ ext{C}^2\),- \(q_1\) and \(q_2\) are the charges,- \(r\) is the distance between the charges.
03

Substitute Values into Coulomb's Law

In our problem, both charges are \(2.0 \times 10^{-8} \text{ C}\), and the distance is \(1 \text{ m}\). Substitute these values into the formula:\[ F = \frac{(8.99 \times 10^9) \times (2.0 \times 10^{-8})^2}{1^2} \].
04

Calculate the Electrostatic Force

Perform the calculation:\[ F = 8.99 \times 10^9 \times (4.0 \times 10^{-16}) \]\[ F = 35.96 \times 10^{-7} \text{ N} \]\[ F = 3.596 \times 10^{-6} \text{ N} \].
05

Determine the Tension in the String

Since the particles are on a smooth, horizontal table and connected by a string, the electrostatic force calculated will be the same as the tension in the string due to Newton's third law. Hence, the tension in the string is \(3.596 \times 10^{-6} \text{ N}\).

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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 one of the foundational principles in electrostatics, describing how charged objects interact. It provides the formula to calculate the electrostatic force between two stationary point charges. The electrostatic force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This relationship is expressed as:\[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\]Here, \(k\) is known as Coulomb's constant and has a very large value of approximately \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\). This constant emphasizes the strength of the electrostatic force.
Coulomb’s Law is a key concept because it allows us to quantify the electric force between two charges. This force is not just a function of the magnitude of the charges, but also how far apart they are. When applying Coulomb's Law, it's crucial to consider the sign and magnitude of the charges:
  • If both charges are like (both positive or both negative), they'll repel each other.
  • If they have opposite signs, they'll attract each other.
Understanding this law helps explain a vast array of phenomena, from why electrons orbit atomic nuclei to how electricity flows through wires.
Electrostatic force
The electrostatic force, also known as Coulombic force, is a fundamental interaction between electrically charged objects. In the context of the two charged particles in the exercise, the electrostatic force is a repulsive force acting along the line joining the charges. This force arises due to the natural tendency of like charges to repel each other.Calculation of the electrostatic force involves using the values provided for the charges and the distance between them, applying Coulomb's Law. In this exercise, both particles have the same charge, \(q = 2.0 \times 10^{-8} \text{ C}\), and are 1 meter apart. Plugging these into the law, we calculate:\[F = \frac{(8.99 \times 10^9) \times (2.0 \times 10^{-8})^2}{1^2} = 3.596 \times 10^{-6} \text{ N}\]This force is mutual, meaning both particles exert this force on each other equally and in opposite directions, per Newton's third law. The magnitude of this force directly determines the tension in the string tying the charges, explaining their equilibrium in the scenario provided.
Tension
In physics, tension refers to the force exerted by a string, rope, or cable when it is pulled tight by forces acting from opposite ends. In the scenario of the two charged particles on the table, the tension in the string is a direct result of the electrostatic force of repulsion between the charges.Since electrostatic forces are mutual and equal, the tension in the string must balance this force to maintain equilibrium. This is essential since there are no other forces acting on the system due to the smooth horizontal surface. Therefore, the tension equals the electrostatic force calculated using Coulomb's Law, which is \(3.596 \times 10^{-6} \text{ N}\).Understanding the concept of tension not only applies to electrostatics but also extends to various physical systems and engineering applications.
For students, appreciating how tension synchronizes with forces like frictionless tables and electrostatic forces is crucial for analyzing and solving physics problems.

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Most popular questions from this chapter

A particle of mass \(1 \mathrm{~g}\) and charge \(2.5 \times 10^{-4} \mathrm{C}\) is released from rest in an electric field of \(1.2 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}\). (a) Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis ? (b) How long will it take for the particle to travel a distance of \(40 \mathrm{~cm} ?\) (c) What will be the speed of the particle after travelling this distance ? (d) How much is the work done by the electric force on the particle during this period?

Three equal charges, \(2.0 \times 10^{-6} \mathrm{C}\) each, are held fixed at the three corners of an equilateral triangle of side \(5 \mathrm{~cm}\). Find the Coulomb force experienced by one of the charges due to the rest two.

A uniform field of \(2 \cdot 0 \mathrm{~N} \mathrm{C}^{-1}\) exists in space in \(x\) -direction. (a) Taking the potential at the origin to be zero, write an expression for the potential at a general point \((x, y, z)\), (b) At which points, the potential is \(25 \mathrm{~V} ?\) (c) If the potential at the origin is taken to be \(100 \mathrm{~V}\), what will be the expression for the potential at a general point ? (d) What will be the potential at the origin if the potential at infinity is taken to be zero ? Is it practical to choose the potential at infinity to be zero ?

The bob of a simple pendulum has a mass of \(40 \mathrm{~g}\) and a positive charge of \(4.0 \times 10^{-6} \mathrm{C}\). It makes 20 oscillations in \(45 \mathrm{~s}\). A vertical electric field pointing upward and of magnitude \(2 \cdot 5 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}\) is switched on. How much time will it now take to complete 20 oscillations ?

Ten positively charged particles are kept fixed on the \(x\) -axis at points \(x=10 \mathrm{~cm}, 20 \mathrm{~cm}, 30 \mathrm{~cm}, \ldots, 100 \mathrm{~cm}\). The first particle has a charge \(1.0 \times 10^{-8} \mathrm{C}\), the second \(8 \times 10^{-8} \mathrm{C}\), the third \(27 \times 10^{-8} \mathrm{C}\) and so on. The tenth particle has a charge \(1000 \times 10^{-8} \mathrm{C}\). Find the magnitude of the electric force acting on a \(1 \mathrm{C}\) charge placed at the origin.

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