/*! 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 44 A point charge \(q_{1}=-4.00 \ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 44

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{~m}, y=0.800 \mathrm{~m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{~m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Short Answer

Expert verified
The magnitude and direction of the net electric field at the origin due to these two point charges can be calculated following these steps. They depend on the specific distances, charges, and direction of each individual electric field.

Step by step solution

01

Determine the Distance from Each Charge to the Origin

Start by determining the distance 'r' from the origin to each charge. For \(q_{1}\), this would be the length of the hypotenuse of a triangle with sides of 0.6m and 0.8m, which can be calculated using Pythagoras's theorem: \(r_{1} = \sqrt{(0.6^2 + 0.8^2)}\). For \(q_{2}\), it is directly on the x-axis so \(r_{2} = 0.6\).
02

Calculate the electric field due to each charge

The electric field 'E' due to a charge 'q' at a distance 'r' can be calculated using Coulomb's Law: \(E = k|q|/r^2\), where 'k' is Coulomb's constant. The directions of the electric fields are directly away from \(q_{2}\) (which is positive), and directly towards \(q_{1}\) (which is negative).
03

Determine the Components of Each Electric Field

To add the electric fields together, determine the x and y components of each. For \(E_{1}\), \(E_{1x} = E_{1} \cos\theta\) and \(E_{1y} = E_{1} \sin\theta\), where \(\theta\) is the angle made with the x axis, found using inverse tan: \(\tan\^-1(0.8/0.6)\). For \(E_{2}\), which is aligned with the x-axis, \(E_{2x} = E_{2}\) and \(E_{2y} = 0\).
04

Calculate the Resultant Electric Field

The net electric fields in the x and y directions can be found by adding the component electric fields: \(E_{x(net)} = E_{1x} + E_{2x}\), and \(E_{y(net)} = E_{1y} + E_{2y}\). The magnitude of the net electric field can then be calculated using Pythagoras's theorem: \(E_{net} = \sqrt{E_{x(net)}^2+E_{y(net)}^2}\). Finally, the direction is found using inverse tan: \(\tan^{-1}(E_{y(net)}/E_{x(net)})\).

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
Understanding how and why charges interact is crucial for delving into the world of electricity and magnetism. At the core of this interaction is Coulomb's Law, which gives us the quantitative ability to calculate the electric force between two stationary point charges. It states that the magnitude of the electrostatic force (\( F \) ) between two point charges is directly proportional to the product of the magnitudes of the charges (\( q_1 \) and 聽\( q_2 \) ) and inversely proportional to the square of the distance (\( r \) ) between them. This essential principle can be expressed as:

\[ F = k \frac{|q_1 \times q_2|}{r^2} \]

where \( k \) is the Coulomb's constant (\( 8.987 \times 10^9 \) Nm2/C2). The direction of the force is along the line connecting the two charges, with like charges repelling each other and unlike charges attracting.
Electric Charge
A foundational component of electric fields and forces is the electric charge. It is a property of particles, such as electrons and protons, that determines their electromagnetic interaction. Charges come in two types, positive and negative, with the convention of protons being positive and electrons negative. When charges are stationary, they exhibit an electric field, influencing other charges around them.
In our exercise, we're dealing with two point charges: \(q_1\) and \(q_2\), measured in coulombs (C), which are the SI unit for electric charge. The negative sign on \(q_1\) indicates it has an excess of electrons, while the positive sign on \(q_2\) signals a deficiency of electrons. Knowing the nature of these charges allows us to predict the direction of the electric fields and forces that they will generate.
Electric Field Components
Breaking down complex vectors into their components simplifies calculations in physics, and this method applies to electric fields as well. An electric field is a vector field surrounding an electric charge, and it exerts force on other charges within the field. When dealing with the net electric field from multiple charges, calculating the horizontal (\( x \) ) and vertical (\( y \) ) components is paramount to finding the resultant electric field.
Using trigonometric functions, we can determine the orientation and magnitude of these components. In the given exercise, we calculate each electric field's components, and then sum them to attain the net electric field. The resulting horizontal and vertical components are then recombined using the Pythagorean theorem to obtain the net electric field magnitude. Accuracy in calculating these components is key for a precise understanding of how forces and fields behave in space, providing clear insights into the nature of electromagnetic interactions.

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

Two thin rods, each with length \(L\) and total charge \(+Q,\) are parallel and separated by a distance \(a .\) The first rod has one end at the origin and its other end on the positive \(y\) -axis. The second rod has its lower end on the positive \(x\) -axis. (a) Explain why the \(y\) -component of the net force on the second rod vanishes. (b) Determine the \(x\) -component of the differential force \(d F_{2}\) exerted on a small portion of the second rod, with length \(d y_{2}\) and position \(y_{2},\) by the first rod. (This requires integrating over differential portions of the first rod, parameterized by \(\left.d y_{1} .\right)\) (c) Determine the net force \(\vec{F}_{2}\) on the second rod by integrating \(d F_{2 x}\) over the second rod. (d) Show that in the limit \(a \gg L\) the force determined in part (c) becomes \(\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{a^{2}} \hat{\imath}\). (e) Determine the external work required to move the second rod from very far away to the position \(x=a\), provided the first rod is held fixed at \(x=0 .\) This describes the potential energy of the original configuration. (f) Suppose \(L=50.0 \mathrm{~cm}, a=10.0 \mathrm{~cm}, Q=10.0 \mu \mathrm{C},\) and \(m=500 \mathrm{~g}\). If the two rods are released from the original configuration, they will fly apart and ultimately achieve a particular relative speed. What is that relative speed?

An insulating rigid rod of length \(2 a\) and negligible mass is attached at its center to a pivot at the origin and is free to rotate in the \(x y\) -plane. A small ball with mass \(M\) and charge \(Q\) is attached to one end of the rod. A second small ball with mass \(M\) and no charge is attached to the other end. A constant electric field \(\vec{E}=-E \hat{\imath}\) is present in the region \(y>0\) while the region \(y<0\) has a vanishing electric field. Define \(\vec{r}\) as the vector that points from the center of the rod to the charged end of the rod, and \(\theta\) as the angle between \(\vec{r}\) and the positive \(x\) -axis. The rod is oriented so that \(\theta=0\) and is given an infinitesimal nudge in the direction of increasing \(\theta\). (a) Write an expression for the vector \(\vec{r}\). (b) Determine the torque \(\vec{\tau}\) about the center of the rod when \(0 \leq \theta \leq \pi\). (c) Determine the torque on the rod about its center when \(\pi \leq \theta \leq 2 \pi\). (d) What is the moment of inertia \(I\) of the system about the \(z\) -axis? (e) The potential energy \(U(\theta)\) is determined by \(\tau=-d U / d \theta .\) Use this equation to write an expression for \(U(\theta)\) over the range \(0 \leq \theta \leq 4 \pi\) using the convention that \(U(0)=0 .\) Make sure that \(U(\theta)\) is continuous. (f) The angular velocity of the rod is \(\omega=\omega(\theta) .\) Using \(\tau=I d^{2} \theta / d t^{2}\) show that the energy \(\frac{1}{2} I \omega^{2}+U(\theta)\) is conserved. (g) Using energy conservation, determine an expression for the angular velocity at the \(n\) th time the positive charge crosses the negative \(y\) -axis.

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is at the origin and point charge \(q_{2}=+3.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=3.00 \mathrm{~cm} .\) Point \(P\) is on the \(y\) -axis at \(y=4.00 \mathrm{~cm} .\) (a) Calculate the electric fields \(\vec{E}_{1}\) and \(\vec{E}_{2}\) at point \(P\) due to the charges \(q_{1}\) and \(q_{2}\). Express your results in terms of unit vectors (see Example 21.6 ). (b) Use the results of part (a) to obtain the resultant field at \(P\), expressed in unit vector form.

Consider an infinite flat sheet with positive charge density \(\sigma\) in which a circular hole of radius \(R\) has been cut out. The sheet lies in the \(x y\) -plane with the origin at the center of the hole. The sheet is parallel to the ground, so that the positive \(z\) -axis describes the "upward" direction. If a particle of mass \(m\) and negative charge \(-q\) sits at rest at the center of the hole and is released, the particle, constrained to the \(z\) -axis, begins to fall. As it drops farther beneath the sheet, the upward electric force increases. For a sufficiently low value of \(m,\) the upward electrical attraction eventually exceeds the particle's weight and the particle will slow, come to a stop, and then rise back to its original position. This sequence of events will repeat indefinitely. (a) What is the electric field at a depth \(\Delta\) beneath the origin along the negative \(z\) -axis? (b) What is the maximum mass \(m_{\max }\) that would prevent the particle from falling indefinitely? (c) If \(m

Two very large parallel sheets are \(5.00 \mathrm{~cm}\) apart. Sheet \(A\) carries a uniform surface charge density of \(-8.80 \mu \mathrm{C} / \mathrm{m}^{2},\) and sheet \(B,\) which is to the right of \(A,\) carries a uniform charge density of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2} .\) Assume that the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) \(4.00 \mathrm{~cm}\) to the right of sheet \(A ;\) (b) \(4.00 \mathrm{~cm}\) to the left of sheet \(A ;\) (c) \(4.00 \mathrm{~cm}\) to the right of sheet \(B\).
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Translational Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Physics of Motion

Read Explanation

Magnetism

Read Explanation

Radiation

Read Explanation
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.