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Chapter 21: Problem 29

(a) What must the charge (sign and magnitude) of a \(1.45-\mathrm{g}\) particle be for it to remain stationary when placed in a downward- directed electric field of magnitude 650 \(\mathrm{N} / \mathrm{C} 2\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitnde to its weight?

Short Answer

Expert verified
(a) The charge must be \(2.18 \times 10^{-5}\) C. (b) The electric field is \(1.03 \times 10^{-7}\) N/C.

Step by step solution

01

Understand the Problem

We are asked to find the charge needed for a particle to remain stationary in an electric field. The electric force must balance the gravitational force. For part (b), we need to find the electric field strength where the electric force equals the weight of a proton.
02

Identify Relevant Formulas

For part (a), the electric force on a charge is given by \( F_e = qE \), and the gravitational force is \( F_g = mg \). For the particle to remain stationary, these forces must be equal: \( qE = mg \). Similarly, for part (b), the electric force \( F_e \) must again be equal to the gravitational force \( F_g \) on the proton.
03

Solve for Charge in Part (a)

Set the forces equal: \( qE = mg \). Solve for \( q \): \( q = \frac{mg}{E} \).Substitute the given values: \( m = 1.45 \times 10^{-3} \) kg, \( E = 650 \) N/C, \( g = 9.8 \) m/s虏.Calculate:\[ q = \frac{1.45 \times 10^{-3} \times 9.8}{650} \approx 2.18 \times 10^{-5} \text{ C} \].The charge must be positive to balance a downward-directed field.
04

Solve for Electric Field in Part (b)

For the proton, \( F_e = mg \).Using the formula \( E = \frac{F}{q} = \frac{mg}{e} \), where \( e \) is the charge of a proton (\( e = 1.60 \times 10^{-19} \) C), and the mass of a proton \( m_p = 1.67 \times 10^{-27} \) kg.Calculate:\[ E = \frac{1.67 \times 10^{-27} \times 9.8}{1.60 \times 10^{-19}} \approx 1.03 \times 10^{-7} \text{ N/C} \].

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Key Concepts

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

Electric Force
The concept of electric force is crucial in the study of electromagnetism. It is the force that charged objects exert on each other. This force can either be attractive or repulsive, depending on the types of charges involved. Opposite charges attract, while like charges repel each other.

To calculate the electric force acting on a charged object in an electric field, we use the formula:

\( F_e = qE \)

Where:
  • \( F_e \) is the electric force.
  • \( q \) represents the charge of the particle.
  • \( E \) is the electric field strength.
If we know the charge of the particle and the strength of the electric field, we can determine how strong the electric force on the particle will be. This principle helps us predict and understand how charged particles will move in various electric fields.
Gravitational Force
Gravitational force is a fundamental force of nature and is responsible for the attraction between masses. On Earth, this force is often referred to as weight. It can be calculated using the formula:

\( F_g = mg \)

Where:
  • \( F_g \) is the gravitational force or weight.
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity, approximately \(9.8\, \text{m/s}^2\) on Earth.
Gravitational force acts downwards towards the center of the Earth. When balancing it with other forces, like the electric force, one can determine if an object will remain at rest or accelerate in a given direction. In the context of our previous exercise, balancing the gravitational and electric forces allows us to solve for the unknown charge or field strength needed to keep an object stationary.
Charge of a Particle
A particle's charge is a measure of its ability to exert or experience electric forces. Charges come in two types: positive and negative. The unit of charge is the Coulomb (C).

In problems involving charges, such as in electric fields, we determine the type and amount of charge required for a given condition using the relationship between electric and gravitational forces. Specifically, for a particle to stay motionless in an electric field, the electric force on it must equal the gravitational force:

\[ qE = mg \]

In this equation:
  • \( q \) is the charge, which we solve for when given other known quantities.
  • The sign of the charge is important; for the particle to remain stationary in a downward pointing electric field, it must have a positive charge to counteract gravity.
Understanding charge is vital in fields like electronics, where controlling and manipulating charges allows for the functioning of circuits and devices.

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Most popular questions from this chapter

A Parallel Universe. Imagine a parallel universe in which the electric force has the same properties as in our universe but there is no gravity. In this parallel universe, the sun carries charge \(Q\) , the earth carries charge \(-Q\) , and the electric attraction between them keeps the earth in orbit. The earth in the parallel universe has the same mass, the same orbital radius, and the same orbital period as in our universe. Calculate the value of \(Q\) . (Consult Appendix F as needed)

Three point charges are arranged on a line. Charge \(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{cm} .\) Charge \(q_{1}\) is at \(x=+2.00 \mathrm{cm} .\) What is \(q_{1}\) (magnitude and sign) if the net force on \(q_{3}\) is zero?

Two tiny balls of mass \(m\) carry equal but opposite charges of magnitude \(q\) . They are tied to the same ceiling hook by light strings of length L. When a horizontal uniform electric field \(E\) is turned on, the balls hang with an angle \(\theta\) between the strings (Fig, 21.46\()\) . (a) Which ball (the right or the left) is positive, and which is negative? (b) Find the angle \(\theta\) between the strings in terms of \(E, q, m,\) and \(g\) . (c) As the electric field is gradually increased in strength, what does your result from part (b) give for the largest possible angle \(\theta ?\)

Two charges, one of 2.50\(\mu \mathrm{C}\) and the other of \(-3.50 \mu \mathrm{C}\) , are placed on the \(x\) -axis, one at the origin and the other at \(x=0.600 \mathrm{m}\) , as shown in Fig. 21.36 . Find the position on the \(x\) -axis where the net force on a small charge \(+q\) would be zero. figure can't copy

Three point charges are arranged along the \(x\) -axis. Charge \(q_{1}=+3.00 \mu \mathrm{C}\) is at the origin, and charge \(q_{2}=-5.00 \mu \mathrm{C}\) is at \(x=0.200 \mathrm{m} .\) Charge \(q_{3}=-8.00 \mu \mathrm{C} .\) Where is \(q_{3}\) located if the net force on \(q_{1}\) is 7.00 \(\mathrm{N}\) in the \(-\mathrm{x}\) -direction?
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