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Two small spheres spaced 20.0 \(\mathrm{cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{N} ?\)

Short Answer

Expert verified
Each sphere has about 2666 excess electrons.

Step by step solution

01

Understanding the Problem

We are given two identical small spheres separated by a distance of 20.0 cm, each with the same charge. The force of repulsion between them is measured to be \(4.57 \times 10^{-21} \mathrm{N}\). We need to calculate the number of excess electrons on each sphere responsible for this force.
02

Coulomb's Law Formula Introduction

Coulomb's Law can be used to determine the force between two charges. The formula is: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] where \(F\) is the force between the charges, \(k\) is Coulomb's constant \(8.99 \times 10^9 \mathrm{Nm^2/C^2}\), \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges (0.20 m in this case). Since the charges are equal, \(q_1 = q_2 = q\).
03

Setting Up the Equation

Since the charges are equal, we can rewrite the formula as: \[ F = \frac{k \cdot q^2}{r^2} \] Here, \(F = 4.57 \times 10^{-21} \mathrm{N}\) and \(r = 0.20 \mathrm{m}\).
04

Solving for Charge \(q\)

Rearrange the equation to solve for the charge \(q\): \[ q^2 = \frac{F \cdot r^2}{k} \] Substituting the known values: \[ q^2 = \frac{4.57 \times 10^{-21} \mathrm{N} \cdot (0.20 \mathrm{m})^2}{8.99 \times 10^9 \mathrm{Nm^2/C^2}} \] Calculate \(q\): \[ q = \sqrt{ \frac{4.57 \times 10^{-21} \times 0.04}{8.99 \times 10^9} } \approx 4.27 \times 10^{-16} \mathrm{C} \]
05

Calculating Excess Electrons

The charge of a single electron is \(-1.602 \times 10^{-19} \mathrm{C}\). To find the number of excess electrons: \[ n = \frac{q}{1.602 \times 10^{-19} \mathrm{C}} \] Substitute the calculated charge: \[ n = \frac{4.27 \times 10^{-16} \mathrm{C}}{1.602 \times 10^{-19} \mathrm{C}} \approx 2666 \] Thus, there are approximately 2666 excess electrons on each sphere.

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter that determines how objects interact electrically. Objects can be positively or negatively charged, depending on whether they have an excess or deficiency of electrons. In this context, an electric charge is a measure of excess electrons present in an object.
For example, when two small spheres have identical charges, it means they either have the same number of excess electrons or the same electron deficit. These charged objects create electric fields, leading to forces of either attraction or repulsion, depending on the nature of the charges.
When working with electric charge, one must remember that it is quantized, meaning it exists in discrete amounts, specifically in multiples of the charge of a single electron:
  • The charge of one electron is approximately -1.602 x 10^{-19}$ C.
  • Charge conservation implies the net charge remains constant in a closed system.
Force of Repulsion
The force of repulsion between two like charges is an essential concept in electromagnetism, explained by Coulomb's Law. This law states that the force between any two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
When two objects have like charges (both positive or both negative), they repel each other. This phenomenon is the force of repulsion.
  • The magnitude of the force of repulsion depends on both the amount of charge on each object and how far apart the objects are.
  • The direction of the force is always such that the objects push away from each other, hence the term repulsion.
  • In this exercise, the measured force of repulsion helps calculate the total charge and eventually determine the number of excess electrons causing this force.
Excess Electrons
Excess electrons on an object signify that it is negatively charged. To determine the number of excess electrons on an object, you must first calculate the total electric charge and then divide that by the charge of a single electron.
This problem involves calculating the number of excess electrons on two spheres. Once you have found the total electric charge on each sphere using Coulomb's Law, divide it by the electron charge (1.602 x 10^{-19} C) to find the number of excess electrons:
  • Excess electrons are the additional electrons that give a sphere its negative charge.
  • In our case, each sphere needs about 2666 excess electrons to create the observed force of repulsion.
  • This calculation is vital for understanding how macro-level forces like repulsion result from micro-level particles like electrons.
Coulomb's Constant
Coulomb's constant is a crucial component of Coulomb's Law, as it allows us to calculate the electric force between charges. The constant has a fixed value k = 8.99 x 10^9 Nm^2/C^2, which explains the strength and behavior of the force through which charged particles interact.
Coulomb's constant makes it possible to determine the force between two charges, so it is always included in such calculations. In practical terms:
  • It converts the unit-less product of charges into Newtons, which is a measure of force.
  • It also suggests how forces scale with distance and charge magnitude.
  • By using Coulomb’s constant in the formula, this exercise seamlessly links the known force with the number of excess electrons.
Understanding Coulomb's constant helps in visualizing the direct relationship between microscopic particles and macroscopic forces in the world around us.

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Most popular questions from this chapter

Excess electrons are placed on a small lead sphere with mass 8.00 g so that its net charge is \(-3.20 \times 10^{-9} \mathrm{C}\) (a) Find the number of excess electrons on the sphere. (b) How many excess electrons are there per lead atom? The atomic number of lead is \(82,\) and its atomic mass is \(207 \mathrm{g} / \mathrm{mol} .\)

Two horizontal, infinite, plane sheets of charge are separated by a distance \(d\) . The lower sheet has negative charge with uniform surface charge density \(-\sigma<0 .\) The upper sheet has positive charge with uniform surface charge density \(\sigma>0 .\) What is the electric field (magnitude, and direction if the field is nonzero) (a) above the upper sheet, (b) below the lower sheet, (c) between the sheets?

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\) . The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\text { max }}\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. ( \(b\) ) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\text { max }}\) and \(d\) if \(E=500 \mathrm{N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{m} / \mathrm{s},\) and \(\alpha=30.0^{\circ} .\)

If Atoms Were Not Neutral... Because the charges on the electron and proton have the same absolute value, atoms are electrically neutral. Suppose this were not precisely true, and the absolute value of the charge of the electron were less than the charge of the proton by 0.00100\(\%\) (a) Estimate what the net charge of this textbook would be under these circumstances. Make any assumptions you feel are justified, but state clearly what they are. (Hint: Most of the atoms in this textbook have equal numbers of electrons, protons, and neutrons.) (b) What would be the magnitude of the electric force between two textbooks placed 5.0 \(\mathrm{m}\) apart? Would this force be attractive or repulsive? Batimate what the acceleration of each book would be if the books were 5.0 \(\mathrm{m}\) apart and there were no nonelectrical forces on them. (c) Discuss how the fact that ordinary matter is stable shows that the absolute values of the charges on the electron and proton must be identical to a very high level of accuracy.

A positive point charge \(q\) is placed at \(x=a,\) and a negative point charge \(-q\) is placed at \(x=-a\) . (a) Find the magnitude and direction of the electric field at \(x=0\) .(b) Derive an expression for the electric field at points on the \(x\) - axis. Use your result to graph the \(x\) -component of the electric field as a function of \(x\) , for values of \(x\) between \(-4 a\) and \(+4 a\) .

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