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.

Ch 21: Electric Charge and Electric Field

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 21: Electric Charge and Electric Field

Problem 37a

Chapter 21, Problem 37a

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

Verified step by step guidance

Understand that the electric field due to a point charge is given by the formula:

E = \frac{k q}{r^{2}}

, where

k

is Coulomb's constant,

q

is the charge, and

r

is the distance from the charge to the point of interest.

Identify that at x = 0, the distance from each charge to the point is

a

. Therefore, the electric field due to each charge at x = 0 can be calculated using the formula:

E = \frac{k q}{a^{2}}

Consider the direction of the electric fields. The charge at x = a will produce an electric field pointing towards the positive x-direction at x = 0, while the charge at x = -a will produce an electric field pointing towards the negative x-direction at x = 0.

Since both charges are equal and the distances are the same, the magnitudes of the electric fields produced by each charge at x = 0 are equal. Therefore, the net electric field at x = 0 is the sum of the magnitudes of these fields, but they are in opposite directions.

Calculate the net electric field at x = 0 by subtracting the electric field due to the charge at x = -a from the electric field due to the charge at x = a. Since they are equal in magnitude, the net electric field will be zero, indicating that the fields cancel each other out.

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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 defined as E = F/q, where F is the force experienced by a test charge q. The direction of the electric field is the direction of the force on a positive test charge.

Superposition Principle

The superposition principle states that the total electric field created by multiple charges is the vector sum of the individual fields produced by each charge. This principle allows us to calculate the net electric field at a point by adding the contributions from each charge, considering both magnitude and direction.

Symmetry in Electric Fields

Symmetry in electric fields refers to the predictable pattern of field lines due to symmetrical charge configurations. In this problem, two identical charges placed symmetrically about the origin create fields that can be analyzed using symmetry, simplifying calculations by recognizing that certain components may cancel out or reinforce each other.

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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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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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