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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?

Short Answer

Expert verified
The force required is \(3.78 \times 10^{18} \text{ N}\), far beyond human capability.

Step by step solution

01

Calculate the Mass of Each Hand

Since each hand is 1% of the person's body weight, and the person weighs 75 kg, we have:\[ \text{Mass of each hand} = 0.01 \times 75 \text{ kg} = 0.75 \text{ kg} \]
02

Calculate the Number of Moles of Calcium in Each Hand

Knowing that the weight of each hand is due to calcium, we use the molar mass of calcium (40.18 g/mol) to find the number of moles:\[ \text{Number of moles} = \frac{0.75 \times 1000}{40.18} \approx 18.67 \text{ moles} \]
03

Calculate the Number of Calcium Atoms in Each Hand

Using Avogadro's number \((6.022 \times 10^{23} \text{ atoms/mol})\), calculate the number of calcium atoms:\[ \text{Number of Ca atoms} = 18.67 \times 6.022 \times 10^{23} \approx 1.124 \times 10^{25} \text{ atoms} \]
04

Calculate the Unbalanced Positive Charge in Each Hand

Each calcium atom has 20 protons. If only 1% of these protons have unbalanced charge, we calculate:\[ \text{Unbalanced charges} = 1.124 \times 10^{25} \times 0.01 \times 20 \approx 2.248 \times 10^{23} \text{ charges} \]Each unbalanced charge corresponds to the charge of one proton: \(1.6 \times 10^{-19} \text{ C}\):\[ \text{Total unbalanced charge} = 2.248 \times 10^{23} \times 1.6 \times 10^{-19} \approx 35968 \text{ C} \]
05

Calculate the Force Between the Hands

Using Coulomb's law for the force between two point charges, \( F = \frac{k \cdot q_1 \cdot q_2}{r^2} \), with \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) and the charges and distance given:\[ F = \frac{8.99 \times 10^9 \cdot 35968^2}{1.7^2} \approx 3.78 \times 10^{18} \text{ N} \]
06

Conclusion on Human Capability to Exert Force

The force calculated is extraordinarily large, \(3.78 \times 10^{18} \text{ N}\), far beyond any human capability. It is not plausible for a human to exert such a force, highlighting the immense force due to even slight imbalances of charge in large numbers of atoms.

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Key Concepts

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

Electrostatic Forces
Electrostatic forces are the forces between charged objects. They are fundamental to many aspects of science and technology. These forces follow Coulomb's Law, which quantifies the attraction or repulsion between two charged particles.

According to Coulomb's Law, the magnitude of the electrostatic force (F) between two point charges (q_1 and q_2) separated by a distance r can be calculated using the formula:
\[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]
where k is the electrostatic constant, approximately 8.99 \times 10^9 \, \mathrm{N \, m^2/C^2}.

Electrostatic forces are extremely strong compared to gravitational forces. This is why even small charge imbalances, such as those calculated in the exercise, can result in enormous forces. Very often, we are unaware of these forces because objects are generally electrically neutral, meaning they have an equal number of positive and negative charges.
Charge Imbalance
A charge imbalance occurs when there is an unequal number of positive charges (protons) and negative charges (electrons) in an object. Such imbalances can result in significant electrostatic forces, as demonstrated in this exercise.

In an atom, electrons usually balance the number of protons, making atoms electrically neutral. However, if the balance is disrupted—even by a minuscule percentage—large electrostatic forces can result.

Specifically, the exercise considers a 1% imbalance in charge, which leads to a massive calculated force between two hands acting as point charges. This scenario illustrates how sensitive electrostatic forces are to even the smallest of charge imbalances.

It is crucial to understand that in reality, such imbalances are rare and typically adjusted swiftly to restore neutrality.
Calcium Atoms
Calcium atoms play a key role in this exercise as they represent the atomic basis for calculating the charge imbalance. Calcium, with the chemical symbol Ca, has the atomic number 20, meaning each calcium atom contains 20 protons and, when neutral, 20 electrons.

The molar mass of calcium is about 40.18 grams per mole, which is used to calculate the number of moles of calcium present in each hand based on its mass.

By knowing the mass and the number of moles, you can determine how many calcium atoms correspond to the mass of a hand, further calculating the number of unbalanced charges. Each unbalanced charge corresponds to a missing electron for one of these protons. In the context of our exercise, this helps assess electrostatic forces develop due to charge imbalances.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry and physics used to express the number of particles, such as atoms or molecules, in a mole of substance. Its value is approximately 6.022 \times 10^{23} \text{ particles/mol}.

In our scenario, we use Avogadro's number to determine the total number of calcium atoms in a hand from the number of moles of calcium:
\[ \text{Number of Ca atoms} = \text{Number of moles} \times 6.022 \times 10^{23} \]

Avogadro's number facilitates conversions between the macroscopic scale of grams and the microscopic scale of individual atoms. Without this constant, it would be challenging to bridge the gap between measurable quantities and atomic-scale phenomena.

In practical applications, this allows scientists to predict how much of a substance is involved in a chemical reaction or a physical process at the atomic level. It is essential for calculating charge imbalances in the context of electrostatic force studies.

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Most popular questions from this chapter

Four identical charges \(Q\) are placed at the corners of a square of side \(L .\) (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

CP Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius \(r=5.29 \times 10^{-11} \mathrm{m}\) around a stationary proton. What is the speed of the electron in its orbit?

CP Operation of an Inkjet Printer. In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of \(1.4 \times 10^{-8}\) g each, leave the nozzle and travel toward the paper at \(20 \mathrm{m} / \mathrm{s},\) passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 \(\mathrm{cm}\) long where there is a uniform vertical electric field with magnitude \(8.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) . If a drop is to be deflected 0.30 \(\mathrm{mm}\) by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop?

CP Two identical spheres are each attached to silk threads of length \(L=0.500 \mathrm{m}\) and hung from a common point (Fig. P21.68). Each sphere has mass \(m=8.00 \mathrm{g} .\) The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_{1},\) and the other a different positive charge \(q_{2} ;\) this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta=20.0^{\circ}\) with the vertical. (a) Draw a free-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the information you have been given, what can you say about the magnitudes of \(q_{1}\) and \(q_{2} ?\) Explain your answers. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of \(30.0^{\circ}\) with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

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.

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