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An amoeba has \(1.00 \times 10^{16}\) protons and a net charge of \(0.300 \mathrm{pC}\). (a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons?

Short Answer

Expert verified
(a) 1.874 million fewer electrons, (b) 1.874 billionths of protons have no electrons.

Step by step solution

01

Understand the concepts

The net charge of an object is the difference between the total charges contributed by protons and electrons. Each proton has a charge of +1 elementary charge, and each electron has a charge of -1 elementary charge. Net charge given in the problem helps determine the difference in the number of protons and electrons.
02

Convert net charge to elementary charges

The net charge is given as 0.300 picocoulombs (pC). First, convert this to coulombs: \[0.300\, \text{pC} = 0.300 \times 10^{-12} \text{C}\]Using the elementary charge, \( e = 1.602 \times 10^{-19} \text{C}\), find how many elementary charges this corresponds to:\[\text{Net charge in elementary charges} = \frac{0.300 \times 10^{-12}}{1.602 \times 10^{-19}}\approx 1.874 \times 10^6\]
03

Calculate the difference in number of protons and electrons

The net amount implies that there are more protons than electrons by the number of elementary charges calculated. So,\[ \text{Fewer electrons than protons} = 1.874 \times 10^6 \]
04

Calculate the fraction of protons without electrons

The problem asks for the fraction of protons without paired electrons. Given that the amoeba has total protons:\[1.00 \times 10^{16}\] and there are \(1.874 \times 10^6\) more protons than electrons:\[ \text{Fraction} = \frac{1.874 \times 10^6}{1.00 \times 10^{16}} \approx 1.874 \times 10^{-10} \]

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Key Concepts

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

Net Charge Calculation
To determine the net charge of an object, you need to look at the difference between positive charges (from protons) and negative charges (from electrons). Protons and electrons hold equal but opposite charges, so when protons and electrons are balanced, their charges effectively cancel each other out, resulting in a net charge of zero.

If there is a difference, the balance shifts, creating a net charge. In the exercise, the amoeba has a net charge of 0.300 pC, which we first convert into coulombs for proper calculations. Once in coulombs, the calculation involves dividing by the elementary charge (the smallest unit of charge) to find the difference in number between protons and electrons. This process provides a clearer picture of where the imbalance lies and lets us determine how many more protons than electrons are in the amoeba.

Understanding the net charge helps explain why objects can exert electrical forces on one another, as the unbalanced charges create an electric field.
Elementary Charge
The elementary charge is a fundamental physical constant denoting the smallest unit of electric charge. It is the charge of a single proton, which is positive, or equivalently, the magnitude of the charge of a single electron, which is negative. The value of the elementary charge is approximately \[ 1.602 \times 10^{-19} \text{C} \]. This all-important constant is critical in calculating the net charge or any electron/proton-related equations.

In calculations where the net charge is given in an unusual unit like picocoulombs (pC), it's essential to convert to coulombs first. Then, dividing the total charge by the elementary charge allows us to understand how many complete units of protons or electrons contribute to that charge.

When performing such calculations, we're essentially counting how many excess protons exist compared to the electrons, giving us an exact number that characterizes the imbalance in charge.
Proton-Electron Balance
In any neutral object, protons and electrons exist in equal numbers, thus cancelling each other's charge perfectly. However, if there is a net charge, like in our given amoeba example, it indicates an imbalance in this delicate proton-electron equilibrium.

To determine how many fewer electrons than protons exist, we use the net charge measured in elementary charges. Our calculations from the net charge give insight into how many electrons are missing compared to protons. This directly tells us how many positive charges are left unpaired, which is the basis for the net charge.

In practical terms, this calculation can also tell us the fraction of protons that lack an electron companion. By knowing the exact number of protons and the disparity in the electron count, calculating the specific fraction is straightforward. This balance of charges helps understand not only basic charge interactions but also more complex behaviors in electrical phenomena.

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Most popular questions from this chapter

(a) What is the electric field \(5.00 \mathrm{~m}\) from the center of the terminal of a Van de Graaff with a \(3.00 \mathrm{mC}\) charge, noting that the field is equivalent to that of a point charge at the center of the terminal? (b) At this distance, what force does the field exert on a \(2.00 \mu \mathrm{C}\) charge on the Van de Graaff's belt?

A test charge of \(+2 \mu \mathrm{C}\) is placed halfway between a charge of \(+6 \mu \mathrm{C}\) and another of \(+4 \mu \mathrm{C}\) separated by \(10 \mathrm{~cm}\). (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the \(+6 \mu \mathrm{C}\) charge \() ?\)

In regions of low humidity, one develops a special "grip" when opening car doors, or touching metal door knobs. This involves placing as much of the hand on the device as possible, not just the ends of one's fingers. Discuss the induced charge and explain why this is done.

Shows an electron passing between two charged metal plates that create an \(100 \mathrm{~N} / \mathrm{C}\) vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) The initial speed of the electron is \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\), and the horizontal distance it travels in the uniform field is \(4.00 \mathrm{~cm}\). (a) What is its vertical deflection? (b) What is the vertical component of its final velocity? (c) At what angle does it exit? Neglect any edge effects.

(a) By what factor must you change the distance between two point charges to change the force between them by a factor of \(10 ?\) (b) Explain how the distance can either increase or decrease by this factor and still cause a factor of 10 change in the force.

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