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An electric field of \(260000 \mathrm{N} / \mathrm{C}\) points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of \(-7.0 \mu \mathrm{C}\) at this spot?

Short Answer

Expert verified
The force is 1.82 N and is directed due east.

Step by step solution

01

Understand the Given Data and Problem

We have an electric field of magnitude \(260000 \, \mathrm{N}/\mathrm{C}\) directed due west. A charge of \(-7.0 \, \mu\mathrm{C}\) is placed in this field. We need to calculate the magnitude and direction of the force on this charge.
02

Convert Microcoulombs to Coulombs

The charge is given as \(-7.0 \, \mu\mathrm{C}\). To use this in our equations, we need to convert it to coulombs. We know that \(1 \, \mu\mathrm{C} = 1 \times 10^{-6} \, \mathrm{C}\). Thus, \(q = -7.0 \, \mu\mathrm{C} = -7.0 \times 10^{-6} \, \mathrm{C}\).
03

Use the Formula for Electric Force

The force \(F\) on a charge \(q\) in an electric field \(E\) is given by the formula:\[ F = qE \]Substitute \(q = -7.0 \times 10^{-6} \, \mathrm{C}\) and \(E = 260000 \, \mathrm{N}/\mathrm{C}\) into the equation.
04

Calculate the Magnitude of the Force

Substituting the values, we have:\[ F = (-7.0 \times 10^{-6} \, \mathrm{C})(260000 \, \mathrm{N}/\mathrm{C}) \]This gives:\[ F = -1.82 \, \mathrm{N} \]The magnitude is \(1.82 \, \mathrm{N}\). The negative sign indicates direction.
05

Determine the Direction of the Force

The negative sign in the force \(-1.82 \, \mathrm{N}\) indicates that the force direction is opposite to that of the electric field. Since the electric field is due west, the force on the negative charge is directed due east.

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Key Concepts

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

Electric Force
Electric force is a fundamental concept in physics, describing the interaction between charged particles. It arises from the influence of an electric field on a charge. In essence, whenever a charged particle, which we often denote as "q," is positioned in an electric field "E," it experiences a force. This force can either be attractive or repulsive, depending on the nature of the charges. The basic formula for calculating the electric force is: \[ F = qE \]. Here, \(F\) is the electric force, \(q\) is the charge, and \(E\) is the electric field. The force is measured in newtons (N), revealing how much push or pull the charge experiences. While applying the equation is straightforward, understanding the underlying principles allows students to predict how charges will behave in various situations, like moving closer or further from each other, or changing the strength of the electric field around them.
Electric Charge
Electric charge is a property of matter that leads to electric force between particles. It is a conserved quantity, meaning that it doesn’t get lost but can be transferred from one place to another. Charges come in two types: positive and negative. The classic example is the charge of a proton (positive) and an electron (negative). Charges are measured in coulombs (C), but in many physical problems, we might see microcoulombs (\mu \mathrm{C}), where \(1\mu\mathrm{C} = 1 \times 10^{-6} \, \mathrm{C}\). Some important things to remember about electric charge include:
  • Like charges repel, opposite charges attract.
  • Charge is quantized, which means it occurs in discrete amounts (multiples of the elementary charge).
  • The net charge of an isolated system remains constant over time.
Understanding electric charge is critical because it is the foundation upon which electric fields and forces are constructed.
Magnitude and Direction of Force
Understanding the magnitude and direction of force in an electric field is crucial for analyzing electromagnetic interactions. The magnitude tells us how strong the force is, while the direction indicates where the force is pushing or pulling. For a charge in an electric field, the force magnitude is calculated by multiplying the charge by the strength of the field: \[ F = qE \] In our example, with the charge of \(-7.0 \, \mu \mathrm{C}\) in a field of \(260000 \, \mathrm{N}/\mathrm{C}\), the magnitude of force is \(1.82 \, \mathrm{N}\). The sign of the charge plays a crucial role in determining direction. A positive charge will align with the field direction, while a negative charge will move opposite. Some key points to remember:
  • Magnitude represents the size or strength of the force.
  • Direction depends on the nature of the charge and the orientation of the field.
  • A negative charge in a westward field will experience force eastward.
By mastering these concepts, students can predict the behavior of charges in various conditions.

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Most popular questions from this chapter

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of \(8480 \mathrm{N} / \mathrm{C}\). The mass of the water drop is \(3.50 \times 10^{-9} \mathrm{kg} .\) (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?

mmh A solid nonconducting sphere has a positive charge \(q\) spread uniformly throughout its volume. The charge density or charge per unit volume, therefore, is \(\frac{q}{\frac{4}{3} \pi R^{3}} .\) Use Gauss' law to show that the electric field at a point within the sphere at a radius \(r\) has a magnitude of \(\frac{q r}{4 \pi \epsilon_{0} R^{3}}\) (Hint: For a Gaussian surface, use a sphere of radius r centered within the solid sphere of radius. Note that the net charge within any volume is the charge density times the volume.)

A cube is located with one corner situated at the origin of an \(x, y,\) \(z\) coordinate system. One of the cube's faces lies in the \(x, y\) plane, another in the \(y, z\) plane, and another in the \(x, z\) plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are \(0.20 \mathrm{m}\) long. A uniform electric field is parallel to the \(x, y\) plane and points in the direction of the \(+y\) axis. The magnitude of the field is \(1500 \mathrm{N} / \mathrm{C}\). (a) Using the outward normal for each face of the cube, find the electric flux through each of the six faces. (b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as Gauss' law predicts, since there is no net charge within the cube.

mmh Two spherical shells have a common center. \(A-1.6 \times 10^{-6}\) C charge is spread uniformly over the inner shell, which has a radius of \(0.050 \mathrm{m} . \mathrm{A}+5.1 \times 10^{-6} \mathrm{C}\) charge is spread uniformly over the outer shell, which has a radius of \(0.15 \mathrm{m}\). Find the magnitude and direction of the electric field at a distance (measured from the common center) of (a) \(0.20 \mathrm{m}\) (b) \(0.10 \mathrm{m},\) and (c) \(0.025 \mathrm{m}\)

An electrically neutral model airplane is flying in a horizontal circle on a 3.0 -m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50.0 J. Reconsider the same situation, except that now there is a point charge of \(+q\) on the plane and a point charge of \(-q\) at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is \(51.8 \mathrm{J}\). Find the magnitude of the charges.

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