/*! 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 31 An electric field of \(260000 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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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 has a magnitude of 1.82 N and points east.

Step by step solution

01

Understand the Given Values

We are given an electric field, \( E = 260000 \, \text{N/C} \), which points due west. We also have a charge, \( q = -7.0 \, \mu C = -7.0 \times 10^{-6} \, \text{C} \). Our task is to find the magnitude and direction of the electric force \( F \) experienced by this charge.
02

Recall the Formula for Electric Force

The electric force \( F \) experienced by a charge \( q \) in an electric field \( E \) is given by the formula: \[ F = qE \] This formula will help us calculate the force using the provided values.
03

Calculate the Magnitude of the Force

Substitute the given values into the formula: \[ F = (-7.0 \times 10^{-6} \, \text{C}) \times (260000 \, \text{N/C}) \] Upon calculation, we find: \[ F = -1.82 \, \text{N} \] The magnitude of the force is \( 1.82 \text{ N} \).
04

Determine the Direction of the Force

Since the charge is negative, the direction of the force \( F \) is opposite to the direction of the electric field. Given that the electric field points west, the force will point east.

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Key Concepts

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

Electric Field
An electric field is a fundamental concept in physics that describes the influence exerted by electric charges. It is a vector field surrounding any electric charge, and it represents the force that would be felt by a positive charge placed within the field. Understanding electric fields is crucial when studying how charged objects interact with one another. The strength or intensity of an electric field is measured in newtons per coulomb (N/C).
  • A strong electric field implies a significant force would act on any charge within it.
  • The direction of the electric field is the direction a positive test charge would move if placed within the field.
This directional nature makes electric fields essential in determining the behavior of charged particles, allowing us to predict the motion based on the field's direction.
Charge
Charge is a property of matter that allows it to experience an electric force. It comes in two types: positive and negative. Atoms are composed of electrons (negative charge) and protons (positive charge), and the imbalance of these subatomic particles gives rise to the net charge. In the context of electric fields, charge is the quantity that determines how strongly a field exerts force on the object.
  • A positive charge within an electric field experiences force in the direction of the field.
  • A negative charge experiences force opposite to the field direction.
Characteristics of charge are central to understanding how electric fields affect the motion of charged particles, such as in the exercise provided where a negative charge is influenced by an electric field pointing west, resulting in a force pointing east.
Force Direction
When it comes to the interaction between a charge and an electric field, the direction of the force is determined by both the charge type and the field's direction. The electric force experienced by a charge is calculated using the formula \( F = qE \), where \( E \) is the electric field and \( q \) is the charge.
  • For a positive charge, the direction of the force is the same as that of the electric field.
  • For a negative charge, the force direction is reversed, meaning it moves opposite to the field.
In practical applications, identifying the force direction helps in predicting how a charged particle will move when situated in an electric field. This understanding clarifies why, in the exercise above, a negative charge experiences force towards the east, contrasting with the westward electric field.

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Most popular questions from this chapter

In a vacuum, two particles have charges of \(q_{1}\) and \(q_{2}\), where \(q_{1}=\) \(+3.5 \mu \mathrm{C} .\) They are separated by a distance of \(0.26 \mathrm{m},\) and particle 1 experiences an attractive force of \(3.4 \mathrm{N} .\) What is \(q_{2}\) (magnitude and \(\operatorname{sign}\) )?

A surface completely surrounds a \(+2.0 \times 10^{-6} \mathrm{C}\) charge. Find the electric flux through this surface when the surface is (a) a sphere with a radius of \(0.50 \mathrm{m},\) (b) a sphere with a radius of \(0.25 \mathrm{m},\) and \((\mathrm{c})\) a cube with edges that are \(0.25 \mathrm{m}\) long.

Two charges are placed between the plates of a parallel plate capacitor. One charge is \(+q_{1}\) and the other is \(q_{2}=+5.00 \mu \mathrm{C}\). The charge per unit area on each of the plates has a magnitude of \(\sigma=1.30 \times 10^{-4} \mathrm{C} / \mathrm{m}^{2} .\) The magnitude of the force on \(q_{1}\) due to \(q_{2}\) equals the magnitude of the force on \(q_{1}\) due to the electric field of the parallel plate capacitor. What is the distance \(r\) between the two charges?

Suppose you want to determine the electric field in a certain region of space. You have a small object of known charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric field. (a) The object has a charge of \(+20.0 \mu \mathrm{C}\) and the instrument indicates that the electric force exerted on it is \(40.0 \mu \mathrm{N}\), due east. What are the magnitude and direction of the electric field? (b) What are the magnitude and direction of the electric field if the object has a charge of \(-10.0 \mu \mathrm{C}\) and the instrument indicates that the force is \(20.0 \mu \mathrm{N},\) due west?

A charge \(+q\) is located at the origin, while an identical charge is located on the \(x\) axis at \(x=+0.50 \mathrm{m} .\) A third charge of \(+2 q\) is located on the \(x\) axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

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