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Ch 21: Electric Charge and Electric Field
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 21, Problem 34b

A +8.75+8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a 6.50-6.50-mC point charge by a light, nonconducting 2.502.50-cm wire. A uniform electric field of magnitude 1.85×1081.85\(\times\)10^8 N/CN/C is directed parallel to the wire, as shown in Fig. E21.3421.34. What would the tension be if both charges were negative?
Two point charges, -6.50 μC and 8.75 μC, connected by a 2.50 cm wire, with an electric field pointing right.

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1
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.
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Coulomb's Law

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.
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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 uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.601.60 cm distant from the first, in a time interval of 3.20×1063.20\(\times\)10^{-6} s. Find the magnitude of the electric field.

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Textbook Question

Two positive point charges qq are placed on the xx-axis, one at x=ax = a and one at x=ax = -a. Derive an expression for the electric field at points on the xx-axis. Use your result to graph the xx-component of the electric field as a function of xx, for values of xx between 4a-4a and +4a+4a.

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Textbook Question

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.601.60 cm distant from the first, in a time interval of 3.20×1063.20\(\times\)10^{-6} s. Find the speed of the proton when it strikes the negatively charged plate.

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Textbook Question

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

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Two positive point charges qq are placed on the xx-axis, one at x=ax = a and one at x=ax = -a. Find the magnitude and direction of the electric field at x=0x = 0.

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Textbook Question

A +8.75+8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a 6.50-6.50-mC point charge by a light, nonconducting 2.502.50-cm wire. A uniform electric field of magnitude 1.85×1081.85\(\times\)10^8 N/CN/C is directed parallel to the wire, as shown in Fig. E21.3421.34. Find the tension in the wire.

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