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Two small spheres spaced 20.0 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 3.33 \(\times\) \(10^{-21}\) N?

Short Answer

Expert verified
Each sphere must have about 537 excess electrons.

Step by step solution

01

Understand Coulomb's Law

Coulomb's Law explains the force between two charged objects. It is given by:\[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the force, \( k \) is the Coulomb's constant \( 8.988 \times 10^9 \text{ N m}^2/\text{C}^2 \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges. Here, \( q_1 = q_2 = q \) and \( r = 0.2 \text{ meters} \).
02

Solve for Charge

Rearrange Coulomb's Law to solve for the charge \( q \):\[ q = \sqrt{\frac{F r^2}{k}} \]Substitute the given \( F = 3.33 \times 10^{-21} \text{ N} \), \( r = 0.2 \text{ m} \), and \( k = 8.988 \times 10^9 \text{ N m}^2/\text{C}^2 \):\[ q = \sqrt{\frac{3.33 \times 10^{-21} \times (0.2)^2}{8.988 \times 10^9}} \approx \sqrt{7.396 \times 10^{-33}} \approx 8.6 \times 10^{-17} \text{ C} \]
03

Determine Number of Electrons

The charge of one electron is approximately \( 1.602 \times 10^{-19} \text{ C} \). To find the number of excess electrons \( n \), use:\[ n = \frac{q}{e} \]where \( e \) is the elementary charge. Therefore,\[ n = \frac{8.6 \times 10^{-17}}{1.602 \times 10^{-19}} \approx 537 \]
04

Interpret the Result

Thus, each sphere must have approximately 537 excess electrons to produce the given force of repulsion.

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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 causes it to experience a force when placed in an electromagnetic field. In simple terms, it's what makes particles attract or repel each other. There are two types of electric charges: positive and negative. These are often carried by protons and electrons, respectively. Charges of the same type repel, while opposite charges attract. This is easy to see in daily life, such as when tiny pieces of paper cling to a charged balloon.
Understanding electric charge is crucial in the context of Coulomb's Law, as it determines how charged objects interact. Coulomb's Law quantifies the force between two charges, allowing us to calculate how much they attract or repel.
In our example, each sphere has an equal and like charge, meaning they will repel each other. The amount of excess charge, in terms of electrons, helps us understand the magnitude of this repulsive force.
Force of Repulsion
The force of repulsion occurs when two objects with similar charges push away from each other. This is described by Coulomb's Law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Simply put, if you increase the charge of either object or decrease the distance between them, the force of repulsion becomes stronger.
  • The equation for this is: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( F \) is the force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance, and \( k \) is Coulomb's constant.
  • In this exercise, we calculated the force of repulsion to be \( 3.33 \times 10^{-21} \text{ N} \).
  • This force pushes the two spheres apart because they both carry the same type of charges (either positive or negative).
Understanding the force of repulsion is crucial for interpreting interactions between charged particles, especially in fields like physics and chemistry.
Elementary Charge
An elementary charge is the smallest unit of electric charge that is considered indivisible in classical physics. It is carried by a single proton or electron, where protons are positive and electrons are negative. The value of an elementary charge is exactly \( 1.602 \times 10^{-19} \text{ C} \).This tiny yet incredibly fundamental constant forms the basis for understanding various physical phenomena, such as how much charge is needed to balance forces in an atom or a molecule.In our example, we calculated how many electrons, which each have one negative elementary charge, would be required to account for the total charge causing the force of repulsion.
  • Given a total charge of \( 8.6 \times 10^{-17} \text{ C} \) per sphere, dividing by the elementary charge of an electron shows how many electrons are needed.
  • The result was approximately 537 excess electrons per sphere, which gives insight into the detailed workings of charge and force in this problem.
Understanding the concept of an elementary charge is essential in fields like chemistry and physics, where the behavior of charged particles is a fundamental aspect.

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Most popular questions from this chapter

In a follow-up experiment, a charge of \(+40\) pC was placed at the center of an artificial flower at the end of a 30-cm long stem. Bees were observed to approach no closer than 15 cm from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \space N/C; (b) 16 \space N/C; (c) 2.7 \times 10^{-10} \space N/C; (d) 4.8 \times 10^{-10} \space N/C.\)

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.

An average human weighs about 650 N. If each of two average humans could carry 1.0 C of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650-N weight?

Two point charges are placed on the \(x\)-axis as follows: Charge \(q_1 = +\)4.00 nC is located at \(x =\) 0.200 m, and charge \(q_2 = +\)5.00 nC is at \(x = -\)0.300 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 nC that is placed at the origin?

Two small spheres with mass \(m =\) 15.0 g are hung by silk threads of length \(L =\) 1.20 m from a common point (Fig. P21.62). When the spheres are given equal quantities of negative charge, so that \(q_1 = q_2 = q\), each thread hangs at \(\theta = 25.0^\circ\) from the vertical. (a) Draw a diagram showing the forces on each sphere. Treat the spheres as point charges. (b) Find the magnitude of \(q\). (c) Both threads are now shortened to length \(L =\) 0.600 m, while the charges \(q_1\) and \(q_2\) remain unchanged. What new angle will each thread make with the vertical? (\(Hint\): This part of the problem can be solved numerically by using trial values for \(\theta\) and adjusting the values of \(\theta\) until a self-consistent answer is obtained.)

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