/*! 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 82 CP Two tiny spheres of mass 6.80... [FREE SOLUTION] | 91影视

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CP 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 \(50.0^{\circ}(\) Fig. .21 .82\()\) . (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?

Short Answer

Expert verified
(a) The right sphere is positive. (b) The field magnitude is approximately 1.12 脳 10鈦 N/C.

Step by step solution

01

Understanding the System

We have two spheres with equal and opposite charges, suspended by strings of equal length, forming an angle of 50掳. When an electric field is applied, it causes a horizontal force on the charges, leading to a symmetrical configuration.
02

Determining Charge Polarity (a)

With the electric field directed to the left, positive charges experience a force in the same direction. Since the spheres remain at rest, and given the angles formed by the strings, the left sphere repels the positive electric field. Thus, the right sphere must have a positive charge.
03

Break Down Forces

Identify forces acting on one sphere: gravitational force downwards \( F_g = mg \), tension in the string \( T \), and electric force \( F_e = qE \). These need to be in equilibrium both horizontally and vertically.
04

Vertical Force Equilibrium

For the vertical component, the tension provides the force to counteract gravity. The vertical component of tension (\( T \cos\theta \)) must equal the gravitational force \( mg \).
05

Horizontal Force Equilibrium

Horizontally, the electric force \( F_e \) is counteracted by the horizontal component of tension (\( T \sin\theta \)). Thus we have \( qE = T \sin\theta \).
06

Calculating Electric Field (b)

First, express the tension using the vertical force equation: \( T = \frac{mg}{\cos\theta} \). Substitute into the horizontal equation: \( qE = \frac{mg}{\cos\theta} \sin\theta \). Solve for electric field \( E \) to find: \( E = \frac{mg \tan\theta}{q} \).
07

Substitute Values and Solve

Use given values, mass \( m = 6.80 \times 10^{-6} \text{ kg} \), charge \( q = 72.0 \times 10^{-9} \text{ C} \), \( \theta = 25掳 \) (half of 50掳 for individual string analysis):\[ E = \frac{6.80 \times 10^{-6} \times 9.8 \times \tan(25掳)}{72.0 \times 10^{-9}} \]Calculate to find \( E \approx 1.12 \times 10^4 \text{ N/C} \).

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Key Concepts

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

Charge Polarity
Charge polarity refers to the characteristic of electric charges that determine their interaction with an electric field. In this exercise, each sphere has an equal magnitude of charge but opposite signs. This means one sphere charges is positive, while the other is negative. The electric field applied is directed towards the left, affecting the forces exerted on the spheres.
When a positive charge is exposed to an electric field, it experiences a force in the direction of the field. Conversely, a negative charge will feel a force opposite to the field direction. In our problem, the electric field pushes the positive charged sphere to the left, while the negative one is drawn to the right.
Understanding this principle allows us to identify the charge polarity of the spheres. Since the position of the spheres remains fixed, the one on the right is repelled by the field, indicating it carries the positive charge.
Equilibrium of Forces
In physics, equilibrium of forces occurs when all forces acting on an object balance perfectly, resulting in no net movement. For the spheres in this exercise, equilibrium is key because it keeps them stationary, despite the presence of multiple forces acting upon them, like tension, electric, and gravitational forces.
In the vertical direction, the gravitational force pulling down is balanced by the upward component of the string tension. Horizontally, the electric force exerted by the field is balanced by the horizontal component of tension.
  • The vertical equilibrium condition states: the tension's vertical component exactly balances the gravitational force.
  • The horizontal equilibrium condition requires the tension's horizontal component to counteract the electric force.
Both conditions must be satisfied to ensure the spheres remain hanging at rest, thereby demonstrating equilibrium.
Electric Force
Electric force is the force experienced by a charged object within an electric field. The magnitude of this force is calculated using the formula: \( F_e = qE \), where \( q \) is the charge and \( E \) is the electric field strength.
In this problem, the electric force pulls the charged spheres in a direction that is dependent on the charge polarity and the external electric field. The positive charge is pulled along with the field direction, while the negative charge is pulled oppositely. This force influences the angle formed by the strings of the spheres since each sphere reacts to these forces differently.
Obtaining the electric force is essential to maintaining equilibrium, as it is balanced precisely by the horizontal component of the tension in the strings.
Gravitational Force
Gravitational force acts on every object with mass, pulling it downward toward the center of the Earth. The calculation of gravitational force is straightforward, given by \( F_g = mg \), where \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, typically \( 9.8 \, \text{m/s}^2 \) on Earth's surface.
For our spheres, this force competes with the tensions in the strings. The vertical component of tension counters the gravitational pull to maintain vertical equilibrium. Without enough tension, the spheres would drop; thus, it plays an essential role in keeping them hanging at rest.
Gravitational force is omnipresent and acts downwards, which dictates that any counteracting force鈥攕uch as tension in the string鈥攎ust be correspondingly structured to prevent rotational movement.

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Most popular questions from this chapter

BI0 Electric Field of Axons. A nerve signal is transmitted through a neuron when an excess of 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. Let us look at a 0.10 -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 -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 sharks 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?

BIO Estimate how many electrons there are in your body. Make any assumptions you feel are necessary, but clearly state what they are. (Hint: Most of the atoms in your body have equal numbers of electrons, protons, and neutrons.) What is the combined charge of all these electrons?

What If We Were Not Neutral? A 75 -kg person holds out his arms so that his hands are 1.7 \(\mathrm{m}\) apart. Typically, a person's hand makes up about 1.0\(\%\) of his or her body weight. For round numbers, we shall assume that all the weight of each hand is due to the calcium in the bones, and we shall treat the hands as point charges. One mole of Ca contains \(40.18 \mathrm{g},\) and each atom has 20 protons and 20 electrons. Suppose that only 1.0\(\%\) of the positive charges in each hand were unbalanced by negative charge. (a) How many Ca atoms does each hand contain? (b) How many coulombs of unbalanced charge does each hand contain? (c) What force would the person's arms have to exert on his hands to prevent them from flying off? Does it seem likely that his arms are capable of exerting such a force?

CALC Two thin rods of length \(L\) lie along the \(x\) -axis, one between \(x=a / 2\) and \(x=a / 2+L\) and the other between \(x=-a / 2\) and \(x=-a / 2-L .\) Each rod has positive charge \(Q\) distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive \(x\) -axis. (b) Show that the magnitude of the force that one rod exerts on the other is $$F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right]$$ (c) Show that if \(a>>L,\) the magnitude of this force reduces to \(F=Q^{2} / 4 \pi \epsilon_{0} a^{2} .\) (Hint: Use the expansion \(\ln (1+z)=z-\) \(z^{2} / 2+z^{3} / 3-\cdots,\) valid for \(|z|<<1 .\) Carry all expansions to at least order \(L^{2} / a^{2} .\) ) Interpret this result.

An average human weighs about 650 \(\mathrm{N} .\) If two such generic humans each carried 1.0 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction hetween them to equal their \(650-\mathrm{N}\) weight?

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