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(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 downwarddirected electric field of magnitude \(650 \mathrm{~N} / \mathrm{C} ?\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Short Answer

Expert verified
Hence, the charge of the stationary particle in the electric field is negative and its magnitude can be calculated with the provided values from Part a. The magnitude of the electric field necessary to balance the weight of a proton is calculated in Part b by substitio the charge and mass of a proton into an adapted form of the electrostatic force formula.

Step by step solution

01

Analyze the situation and determine required variables

From the problem, we know that the particle in part a is stationary despite the presence of an electric field, which suggests that the electrostatic force is balanced by the gravitational force. Therefore, we can create the equation \(|F_{E}| = |F_{G}| \). The force due to gravity can be calculated by multiplying the mass of the particle by the acceleration due to gravity (\(F_{G}= mg\)). The force due to the electric field is calculated by multiplying the electric field (\(E\)) by the charge of a particle (\(q\)). So, \(|F_{E}| = |F_{G}| \) simplifies to \(|qE| = |mg|\)
02

Solve For the Charge

Rearrange our equation \(|qE| = |mg|\) to solve for \(q\). This gives us \(q = mg/E\). Substituting all known values into the equation gives \(q = (1.45 \mathrm{~g} * 9.81 \mathrm{~m/s^2}) / 650 \mathrm{~N/C}\). Make sure to convert the mass from grams to kilograms: \(1.45 \mathrm{~g} = 1.45 * 10^{-3} \mathrm{~kg}\). Perform the multiplication and division to find the charge. Since the gravitational force and the electric force are opposite in direction, the charge \(q\) must be negative.
03

Analyze and setup equation for Part B

Part b asks for the electric field that would balance the weight of a proton. We can continue using the equation \(|F_{E}| = |F_{G}|\) but now use the mass and charge of a proton. The charge of a proton \(q\) is \(1.6 * 10^{-19} \mathrm{C}\) and its mass \(m\) is \(1.67 * 10^{-27} \mathrm{~kg}\).
04

Solve for The Electric Field

Rearrange our equation \(|qE| = |mg|\) to solve for \(E\). This gives us \(E = mg/q\). Substituting all known values into the equation gives \(E = (1.67 * 10^{-27} \mathrm{~kg} * 9.81 \mathrm{~m/s^2}) / 1.6 * 10^{-19} \mathrm{C}\). Performing the multiplication and division yields the magnitude of the required electric field.

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Key Concepts

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

Electric Force
The electric force is a fundamental interaction between objects that possess an electric charge. According to Coulomb's law, the electric force (\( F_E \)) between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is given by \( F_E = k \frac{|q_1 * q_2|}{r^2} \) where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.

In our exercise, the electric force is considered acting on a single charge in an electric field, which can be derived from Coulomb's law and defined as \( F_E = qE \) where \( q \) is the charge of the particle, and \( E \) is the electric field strength. The direction of this force depends on the sign of the charge: it's in the direction of the field for a positive charge and opposite for a negative charge.
Gravitational Force
The gravitational force is another fundamental interaction and it's always an attraction between two masses. Unlike the electric force, which can attract or repel, gravity only pulls objects together. Newton's law of universal gravitation states that every point mass attracts every other point mass by a force acting along the line joining both points. This force is proportional to the product of their masses and inversely proportional to the square of the distance between them. The equation for gravitational force (\( F_G \) is \( F_G = G \frac{m_1 * m_2}{r^2} \) where \( G \) is the gravitational constant.

When considering objects near the Earth's surface, the gravitational force can be simplified to \( F_G = mg \) where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. The exercise uses this simplified formula, equating it to the electric force to find the balance for a stationary object in an electric field.
Charge of a Particle
The charge of a particle is an intrinsic property that determines its electromagnetic interactions. Charges can be positive or negative, and like charges repel each other, while unlike charges attract. The unit of electric charge is the coulomb (\( C \) which is a relatively large unit; subatomic particles like protons and electrons carry a charge of approximately \( \pm1.6 \times 10^{-19} C \).

In the context of the given exercise, calculating the charge, \( q \), necessary for a particle to remain stationary in an electric field, involves balancing the electric force with the gravitational force. The sign of the charge (positive or negative) will determine the direction of the electric force exerted on the particle. For a negatively charged particle, the electric force is in the direction opposite to the gravitational force, which allows the particle to remain stationary when these two forces are equal.
Electrostatics
Electrostatics is the study of electric charges at rest. It encompasses the behaviors of forces between charges that are not moving. Key concepts include electric charge, electric field, electric potential, and the behavior of conductors and insulators. An electric field (\( E \) is a vector field that represents the influence a charge exerts on other charges around it, and is defined as the force per unit charge. It is measured in newtons per coulomb (\( N/C \) or volts per meter (\( V/m \) and can be visualized with field lines emanating from positive charges and ending on negative charges.

The core principle we're applying in the exercise is that a stationary charged particle in an electric field is experiencing no net force, which means the electrostatic force is being exactly balanced by another force, in this case, gravity. Solving problems in electrostatics often involves understanding and manipulating the relationships between charge, electric field, and force.

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Most popular questions from this chapter

A thin disk with a circular hole at its center, called an \(a n-\) nulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. \(\mathbf{P 2 1 . 8 7}\) ). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z\) plane, with its center at the origin. For an arbitrary point on the \(x\) -axis (the axis of the annulus), find the magnitude and direction of the electric field \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus. (c) Show that at points on the \(x\) -axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\) -axis (but cannot move off the axis). The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

Two tiny spheres of mass \(6.80 \mathrm{mg}\) carry charges of equal magnitude, \(72.0 \mathrm{nC}\), but opposite sign. They are tied to the same ceiling hook by light strings of length \(0.530 \mathrm{~m}\). When a horizontal uniform electric field \(E\) that is directed to the left is turned on, the spheres hang at rest with the angle \(\theta\) between the strings equal to \(58.0^{\circ}\) (Fig. P21.74). (a) Which ball (the one on the right or the one on the left) has positive charge? (b) What is the magnitude \(E\) of the field?

Two point charges are placed on the \(x\) -axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{~m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{~m} .\) What are the magnitude and direction of the total force exerted by these two charges on a negative point charge \(q_{3}=-6.00 \mathrm{nC}\) that is placed at the origin?

A nerve signal is transmitted through a neuron when an excess of \(\mathrm{Na}^{+}\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately \(10.0 \mu \mathrm{m}\) in diameter, and measurements show that about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions per meter (each of charge \(+e\) ) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the axon. Consider a \(0.10 \mathrm{~mm}\) length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a \(0.10 \mathrm{~mm}\) length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is \(5.00 \mathrm{~cm}\) below the skin? (c) Certain shark can respond to electric fields as weak as \(1.0 \mu \mathrm{N} / \mathrm{C}\). How far from this segment of axon could a shark be and still detect its electric field?

A straight, nonconducting plastic wire \(8.50 \mathrm{~cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point \(6.00 \mathrm{~cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point \(6.00 \mathrm{~cm}\) directly above its center.

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