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Chapter 18: Problem 37

Interactive Solution \(\underline{18.37}\) at provides a model for problems of this kind. A small object has a mass of \(3.0 \times 10^{-3} \mathrm{~kg}\) and a charge of \(-34 \mu \mathrm{C}\). It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of \(2.5 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\) in the direction of the \(+x\) axis. Determine the magnitude and direction of the electric field.

Short Answer

Expert verified
Electric field magnitude is approximately 220,588 N/C, directed in the -x direction.

Step by step solution

01

Understand the Problem

We are given a mass, charge, acceleration, and need to find the electric field. The object experiences acceleration due to the force exerted by the electric field. The direction of the electric field will be opposite to the charge's direction if the charge is negative.
02

Use Newton's Second Law

According to Newton's Second Law, the force on the object can be calculated as: \[ F = ma \]where \( m = 3.0 \times 10^{-3} \; \text{kg} \) and \( a = 2.5 \times 10^3 \; \text{m/s}^2 \). Substituting the values:\[ F = 3.0 \times 10^{-3} \times 2.5 \times 10^3 \]
03

Calculate the Force

Now, we calculate the force:\[ F = 3.0 \times 10^{-3} \times 2.5 \times 10^3 = 7.5 \; \text{N} \]
04

Relate Force and Electric Field

The electric force \( F \) acting on the charge can also be expressed in terms of electric field \( E \):\[ F = qE \]where \( q = -34 \times 10^{-6} \; \text{C} \). Isolating \( E \), we have:\[ E = \frac{F}{q} \]
05

Calculate the Magnitude of the Electric Field

Substitute the values of force and charge into the electric field equation:\[ E = \frac{7.5}{-34 \times 10^{-6}} \approx -220588 \; \text{N/C} \]The negative sign indicates the direction is opposite to the positive x-axis.
06

Consider the Direction of the Electric Field

Since the charge is negative and the acceleration is in the +x direction, according to physics, the electric field must be in the -x direction to result in the observed acceleration direction for a negative charge.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a central principle in physics that helps us understand how objects move. It tells us that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In a formula, we express it as: \[ F = ma \]
  • Force \( F \): This is measured in Newtons (N). In our problem, the force acting on the small object is derived based on its given mass and acceleration.
  • Mass \( m \): This is the weight of the object. Here, it's \( 3.0 \times 10^{-3} \; \text{kg} \).
  • Acceleration \( a \): This shows how quickly the object's velocity is changing, given as \( 2.5 \times 10^3 \; \text{m/s}^2 \).
By plugging in the values, we find the force needed for the object to accelerate at the given rate. This foundation helps us later connect the force to the electric field, illustrating the interplay between physical laws and electric phenomena.
Electric Charge
Electric charge is a fundamental property of matter that leads to electromagnetic interactions. Charges come in two types: positive and negative. In this problem, the object has a negative charge of \(-34 \; \mu \text{C} \) (microcoulombs), representing the object's total charge.
  • Notation: The value \(-34 \mu \text{C} \) can be transformed into coulombs (C) by recognizing that microcoulombs are \(10^{-6}\) of a coulomb, so the charge is really \( -34 \times 10^{-6} \; \text{C} \).
  • Impact on Direction: Because the charge is negative, it reverses the direction of the electric field. This means we'll see forces act in directions that might initially seem counterintuitive, such as the electric field being opposite the acceleration.
The characteristics of the charge are crucial, as they influence the behavior of the object within an electric field. Knowing the sign and magnitude of the charge allows us to ascertain how the electric field affects the motion of the object.
Acceleration
Acceleration is a measure of how fast an object's velocity changes over time. In physics, it's a vector quantity, meaning it has both a magnitude (size) and a direction. Here, the object experiences an acceleration of \( 2.5 \times 10^3 \; \text{m/s}^2 \) directed along the \(+x\) axis.
  • Magnitude: Shows the speed increase rate, indicated by the high magnitude of \( 2.5 \times 10^3 \; \text{m/s}^2 \). High acceleration means the object's velocity is changing rapidly.
  • Direction: The positive \(x\) direction suggests the object is speeding up to the right on a standard axial coordinate system.
In the context of this exercise, the acceleration is due to the force exerted by the electric field. Since the object is negatively charged, the observed acceleration correlates with the underlying physics principles, where the force direction opposes that of the electric field. This helps us decode the actual direction of the electric field responsible for the motion.

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Most popular questions from this chapter

Four identical metallic objects carry the following charges: \(+1.6,+6.2,-4.8,\) and \(-9.4 \mu \mathrm{C} .\) The objects are brought simultaneously into contact, so that each touches the others. Then they are separated, (a) What is the final charge on each object? (b) How many electrons (or protons) make up the final charge on each object?

Two particles, with identical positive charges and a separation of \(2.60 \times 10^{-2} \mathrm{~m}\), are released from rest. Immediately after the release, particle 1 has an acceleration \(\overrightarrow{\mathbf{a}},\) whose magnitude is \(4.60 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\), while particle 2 has an acceleration \(\overrightarrow{\mathbf{a}}_{2}\) whose magnitude is \(8.50 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\). Particle 1 has a mass $$ \text { of } 6.00 \times 10^{-6} \mathrm{~kg} . \text { Find } $$ (a) the charge on each particle and (b) the mass of particle 2 .

Two charges are located along the \(x\) axis: \(q_{1}=+6.0 \mu \mathrm{C}\) at \(x_{1}=+4.0 \mathrm{~cm}\), and \(q_{2}=+6.0 \mu \mathrm{C}\) at \(x_{2}=-4.0 \mathrm{~cm}\). Two other charges are located on the \(y\) axis: \(q_{3}=+3.0 \mu \mathrm{C}\) at \(y_{3}=+5.0 \mathrm{~cm}\), and \(q_{4}=-8.0 \mu \mathrm{C}\) at \(y_{4}=+7.0 \mathrm{~cm} .\) Find the net electric field (magnitude and direction) at the origin.

A rectangle has a length of \(2 d\) and a height of \(d\). Each of the following three charges is located at a corner of the rectangle: \(+q_{1}\) (upper left corner), \(+q_{2}\) (lower right corner), and \(-q\) (lower left corner). The net electric field at the (empty) upper right corner is zero. Find the magnitudes of \(q_{1}\) and \(q_{2}\). Express your an swers in terms of \(q\).

Two identical small insulating balls are suspended by separate \(0.25-\mathrm{m}\) threads that are attached to a common point on the ceiling. Each ball has a mass of \(8.0 \times 10^{-4} \mathrm{~kg}\). Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of \(36^{\circ}\) between the threads. Determine (a) the charge on each ball and (b) the tension in the threads.
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