Ch 21: Electric Charge and Electric Field

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 21: Electric Charge and Electric Field

Problem 26b

Chapter 21, Problem 26b

A $+8.75$ -mC point charge is glued down on a horizontal frictionless table. It is tied to a $-6.50$ -mC point charge by a light, nonconducting $2.50$ -cm wire. A uniform electric field of magnitude $1.85$\times$10^8$ $N/C$ is directed parallel to the wire, as shown in Fig. E $21.34$ . What would the tension be if both charges were negative?

Verified step by step guidance

First, understand that the tension in the wire is due to the forces acting on the charges. These forces include the electric force due to the electric field and the electrostatic force between the charges.

Calculate the force on each charge due to the electric field using the formula: $ F = qE $, where $ q $ is the charge and $ E $ is the electric field. Since both charges are negative, the direction of the force will be opposite to the direction of the electric field.

Calculate the electrostatic force between the two charges using Coulomb's law: $ F = \frac{k |q_1 q_2|}{r^2} $, where $ k $ is Coulomb's constant, $ q_1 $ and $ q_2 $ are the charges, and $ r $ is the distance between them. Since both charges are negative, the force will be attractive.

Determine the net force on each charge by considering the direction of the forces. The tension in the wire will be equal to the net force acting on either charge, as the system is in equilibrium.

Finally, express the tension in terms of the calculated forces. The tension will be the sum of the magnitudes of the electric force and the electrostatic force, considering their directions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Coulomb's Law

Coulomb's Law describes the electrostatic interaction between electrically charged particles. The force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. This force is attractive if the charges are of opposite signs and repulsive if they are of the same sign.

Electric Field

An electric field is a vector field around a charged particle that represents the force exerted on other charges in the vicinity. The strength and direction of the electric field are defined by the force per unit charge experienced by a positive test charge placed in the field. In this problem, the uniform electric field affects the forces on the charges, influencing the tension in the wire.

Tension in a Wire

Tension in a wire is the force exerted along the wire, which arises due to the forces acting on the objects connected by the wire. In this scenario, the tension results from the balance of electrostatic forces between the charges and the force due to the external electric field. The tension will change if the charges are both negative, as the direction and magnitude of forces will differ.

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Related Practice

Textbook Question

A point charge is placed at each corner of a square with side length $a$ . All charges have magnitude $q$ . Two of the charges are positive and two are negative (Fig. E $21.42$ ). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of $q$ and $a$ ?

Textbook Question

A proton is traveling horizontally to the right at $4.50$\times$10^6$ m/s. How much time does it take the proton to stop after entering the field?

Textbook Question

A $negative 4.00$ -nC point charge is at the origin, and a second $negative 5.00$ -nC point charge is on the $x$ -axis at $x equals 0.800$ m. Find the electric field (magnitude and direction) at each of the following points on the $x$ -axis: (i) $x equals 0.200$ m; (ii) $x equals 1.20$ m; (iii) $x = - 0.200$ m.

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Textbook Question

A proton is traveling horizontally to the right at $4.50$\times$10^6$ m/s. What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)? Note: Part (a) asks for how much time does it take the proton to stop after entering the field.

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Textbook Question

A $positive 8.75$ -mC point charge is glued down on a horizontal frictionless table. It is tied to a $negative 6.50$ -mC point charge by a light, nonconducting $2.50$ -cm wire. A uniform electric field of magnitude $1.85 \times 10^{8}$ $N / C$ is directed parallel to the wire, as shown in Fig. E $21.34$ . Find the tension in the wire.

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Textbook Question

A $negative 4.00$ -nC point charge is at the origin, and a second $-5.00$ -nC point charge is on the $x$ -axis at $x = 0.800$ m. Find the net electric force that the two charges would exert on an electron placed at each point in part (a). Note: Part (a) asked to find the electric field (magnitude and direction) at each of the following points on the $x$ -axis: (i) $x = 0.200$ m; (ii) $x = 1.20$ m; (iii) $x = -0.200$ m.

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