/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A -3.00 nC point charge is on th... [FREE SOLUTION] | 91影视

91影视

A -3.00 nC point charge is on the \(x\) -axis at \(x=1.20 \mathrm{~m}\). A second point charge, \(Q,\) is on the \(x\) -axis at \(-0.600 \mathrm{~m}\). What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) \(45.0 \mathrm{~N} / \mathrm{C}\) in the \(+x\) -direction, (b) \(45.0 \mathrm{~N} / \mathrm{C}\) in the \(-x\) -direction?

Short Answer

Expert verified
The sign and magnitude of Q for the resultant electric field at the origin to be (a) 45.0 N/C in the +x -direction is +2.00 nC, (b) 45.0 N/C in the -x -direction is -2.00 nC.

Step by step solution

01

Calculate the electric field due to the fixed charge

At the origin, the electric field due to the fixed charge at x = 1.20 m, E1 = k|q1|/r鈧伮, where q1 = -3.00 nC, r鈧 = 1.20 m. Substituting the values, E1 = (9 x 10^9 N.m虏/C虏 * 3 x 10^-9 C) / (1.2 m)虏 = -15 N/C in the -x direction, as field due to a negative charge is towards the charge.
02

Calculate the electric field due to the second charge for the resultant field to be in +x direction

For the resultant field at the origin to be 45.0 N/C in the +x direction, the electric field due to the second charge, E2 = E - E1, where E is the resultant field. Substituting the values, E2 = 45.0 N/C - (-15 N/C) = 60 N/C in the +x direction. Now, using the formula for electric field, E = k|q|/r虏, the charge Q = E2.r鈧偮/k. Substituting the values, Q = (60 N/C * (0.6 m)虏) / (9 x 10^9 N.m虏/C虏) = 2.00 nC. Since E2 is in the +x direction, Q must be positive.
03

Calculate the electric field due to the second charge for the resultant field to be in -x direction

For the resultant field at the origin to be 45.0 N/C in the -x direction, the electric field due to the second charge, E2 = E1 - E, where E is the resultant field. Substituting the values, E2 = (-15 N/C) - 45.0 N/C = -60 N/C in the -x direction. Now, using the formula for electric field, E = k|q|/r虏, the charge Q = E2.r鈧偮/k. Substituting the values, Q = (-60 N/C * (0.6 m)虏) / (9 x 10^9 N.m虏/C虏) = -2.00 nC. Since E2 is in the -x direction, Q must be negative.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's law is the cornerstone of electrostatics, illustrating how electric charges interact with each other. It states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula for Coulomb's law is given by
\( F = k \frac{|q_1q_2|}{r^2} \),
where \( F \) is the force between the charges, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance separating them, and \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \) N.m虏/C虏.
Coulomb's law applies to point charges - idealized charges located at a single point in space. Despite this simplification, Coulomb's law provides a fundamental understanding of how charges exert forces on each other and is the basis for calculating electric fields, as seen in the exercise provided.
Electric Force
The electric force is the force experienced by a charge in the presence of another charge or an electric field. When charges are stationary, this is referred to as the electrostatic force, as per Coulomb's law. The direction of the force depends on the nature of the charges; like charges repel each other, while opposite charges attract.

Application in Exercise

In the exercise scenario, the force is determined by the electric field created by the charges. The electric field, denoted as \( E \), at a point in space is defined as the force per unit charge experienced by a small positive test charge placed at that point. It has both magnitude and direction, and for a point charge, it is calculated by rearranging Coulomb鈥檚 law to
\( E = k \frac{|q|}{r^2} \).
In the context of the exercise, the electric field due to the point charge is used to determine the necessary characteristics of a second charge in order to achieve a specific resultant electric field at the origin.
Superposition Principle
The superposition principle is a fundamental concept in physics, particularly in the realm of electrostatics. It posits that when multiple charges are present, the total electric force at a specific point is the vector sum of the electric forces produced by each individual charge acting independently. This principle allows us to calculate complex fields by breaking them down into simpler components.

Application in Exercise

The principle is applied in the exercise by adding the electric fields due to individual charges when evaluating the resultant electric field at the origin. Each charge contributes to the total field, and their individual fields are algebraically added, considering both magnitude and direction, to find the combined effect. Thus, the second charge in the exercise must be chosen in such a way that it creates an electric field that, when combined with the field from the initial charge, results in the desired net electric field at the origin.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Imagine two \(1.0 \mathrm{~g}\) bags of protons, one at the earth's north pole and the other at the south pole. (a) How many protons are in each bag? (b) Calculate the gravitational attraction and the electric repulsion that each bag exerts on the other. (c) Are the forces in part (b) large enough for you to feel if you were holding one of the bags?

A semicircle of radius \(a\) is in the first and second quadrants, with the center of curvature at the origin. Positive charge \(+Q\) is distributed uniformly around the left half of the semicircle, and negative charge \(-Q\) is distributed uniformly around the right half of the semicircle (Fig. P21.84). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge?

A particle has charge \(-5.00 \mathrm{nC}\). (a) Find the magnitude and direction of the electric field due to this particle at a point \(0.250 \mathrm{~m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of \(12.0 \mathrm{~N} / \mathrm{C} ?\)

A small \(12.3 \mathrm{~g}\) plastic ball is tied to a very light \(28.6 \mathrm{~cm}\) string that is attached to the vertical wall of a room (Fig. \(\mathbf{P} 2 \mathbf{1 . 6 3}\) ). A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of \(-1.11 \mu \mathrm{C},\) you observe that it remains suspended, with the string making an angle of \(17.4^{\circ}\) with the wall. Find the magnitude and direction of the electric field in the room.

A very long line of charge with charge per unit length \(+8.00 \mu \mathrm{C} / \mathrm{m}\) is on the \(x\) -axis and its midpoint is at \(x=0 .\) A second very long line of charge with charge per length \(-4.00 \mu \mathrm{C} / \mathrm{m}\) is parallel to the \(x\) -axis at \(y=10.0 \mathrm{~cm}\) and its midpoint is also at \(x=0 .\) At what point on the \(y\) -axis is the resultant electric field of the two lines of charge equal to zero?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.