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 36b

Chapter 21, Problem 36b

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.

Verified step by step guidance

To find the electric field at a point due to a point charge, use the formula: E = (k * |q|) / r^2, where E is the electric field, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point.

For point (i) x = 0.200 m, calculate the electric field due to each charge separately. For q1 at the origin, r = 0.200 m. For q2 at x = 0.800 m, r = 0.600 m. Determine the direction of each field: both charges are negative, so the electric field points towards the charges.

For point (ii) x = 1.20 m, calculate the electric field due to each charge. For q1, r = 1.20 m. For q2, r = 0.400 m. Again, determine the direction of each field: both fields point towards the charges.

For point (iii) x = -0.200 m, calculate the electric field due to each charge. For q1, r = 0.200 m. For q2, r = 1.000 m. Determine the direction of each field: both fields point towards the charges.

To find the net electric force on an electron at each point, use F = e * E, where e is the charge of an electron (1.60 x 10^-19 C). Calculate the force using the net electric field found at each point.

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Key Concepts

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

Electric Field

The electric field is a vector field around a charged particle that represents the force exerted per unit charge at any point in space. It is calculated using Coulomb's law, where the electric field E due to a point charge q at a distance r is given by E = k * |q| / r^2, with k being Coulomb's constant. The direction of the field is radially outward for positive charges and inward for negative charges.

Superposition Principle

The superposition principle states that the net electric field due to multiple charges is the vector sum of the electric fields produced by each charge individually. This principle allows us to calculate the total electric field at a point by considering the contribution from each charge separately and then adding them together, taking into account both magnitude and direction.

Electric Force on a Charge

The electric force on a charge in an electric field is given by F = qE, where F is the force, q is the charge, and E is the electric field at the location of the charge. For an electron, which has a negative charge, the direction of the force is opposite to the direction of the electric field. This concept is crucial for determining the net force on a charge placed in the vicinity of other charges.

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Related Practice

Textbook Question

A point charge $q_{1} = - 4.00$ nC is at the point $x equals 0.600$ m, $y equals 0.800$ m, and a second point charge $q sub 2 equals positive 6.00$ nC is at the point $x equals 0.600$ m, $y equals 0$ . Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

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 $negative 4.00$ -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 $+8.75$ -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?

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Textbook Question

A very long, straight wire has charge per unit length $3.20 \times 10^{- 10}$ C/m. At what distance from the wire is the electric field magnitude equal to $2.50$ N/C?

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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