A point charge q1=−4.00q_1=-4.00 nC is at the point x=0.600x = 0.600 m, y=0.800y = 0.800 m, and a second point charge q2=+6.00q_2=+6.00 nC is at the point x=0.600x = 0.600 m, y=0y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

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

All textbooks

Young & Freedman Calc 14th Edition

Ch 21: Electric Charge and Electric Field

Problem 48

Chapter 21, Problem 48

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.

Verified step by step guidance

Identify the positions of the charges: q1 = -4.00 nC is at (0.600 m, 0.800 m) and q2 = +6.00 nC is at (0.600 m, 0). The origin is at (0, 0).

Calculate the distance from each charge to the origin using the distance formula: $ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $. For q1, the distance is $ r_1 = \sqrt{(0.600)^2 + (0.800)^2} $. For q2, the distance is $ r_2 = 0.600 $ m.

Use Coulomb's Law to find the electric field due to each charge at the origin. The electric field $ E $ due to a point charge is given by $ E = \frac{k |q|}{r^2} $, where $ k $ is Coulomb's constant $ 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 $. Calculate $ E_1 $ for q1 and $ E_2 $ for q2.

Determine the direction of each electric field vector. The electric field due to a negative charge points towards the charge, while the field due to a positive charge points away. For q1, the field points towards (0.600 m, 0.800 m), and for q2, it points away from (0.600 m, 0).

Calculate the net electric field at the origin by vector addition of $ E_1 $ and $ E_2 $. Break each electric field into components, sum the components, and use the Pythagorean theorem to find the magnitude of the resultant vector. Use trigonometry to find the direction of the net electric field.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

17m

Was this helpful?

Key Concepts

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

Electric Field Due to a Point Charge

The electric field (E) created by a point charge (q) at a distance (r) is given by Coulomb's law: E = k|q|/r², where k is Coulomb's constant. The direction of the field is radially outward from a positive charge and inward towards a negative charge. This concept helps determine the individual electric fields at the origin due to each charge.

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 independently. This principle is crucial for calculating the net electric field at the origin by adding the contributions from both q1 and q2, considering their magnitudes and directions.

Vector Addition

Vector addition involves combining vectors by adding their corresponding components. For electric fields, this means summing the x and y components separately to find the resultant vector. This concept is essential for determining the magnitude and direction of the net electric field at the origin from the individual fields of q1 and q2.

Recommended video:

Guided course

07:30

Vector Addition By Components

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

Point charges $q_{1} = - 4.5$ nC and $q sub 2 equals positive 4.5$ nC are separated by $3.1$ mm, forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle of $36.9$ ° with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude $7.2 \times 10^{- 9}$ Nm?

views

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.

views

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 charge of $negative 6.50$ nC is spread uniformly over the surface of one face of a nonconducting disk of radius $1.25$ cm. Why is the field in part (a) stronger than the field in part (b)? Why is the field in part (c) the strongest of the three fields? Note: Part (a) asked to find the magnitude and direction of the electric field this disk produces at a point $P$ on the axis of the disk a distance of $2.00$ cm from its center. Part (b) asked to find the magnitude and direction of the electric field at point $P$ , supposing that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Part (c) asked to find the magnitude and direction of the electric field at point $P$ if the charge is all brought to the center of the disk.

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.

views