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Chapter 17: Problem 10

\(\bullet\) A negative charge of \(-0.550 \mu C\) exerts an upward 0.200 \(\mathrm{N}\) force on an unknown charge 0.300 \(\mathrm{m}\) directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the - 0.550\(\mu \mathrm{C}\) charge?

Short Answer

Expert verified
(a) The unknown charge is approximately +1.21 脳 10鈦烩伒 C. (b) The force is 0.200 N, directed downward.

Step by step solution

01

Understand the given data

We are given that a negative charge \(-0.550 \mu C\) exerts a force of \(0.200 \mathrm{N}\) upward on another unknown charge 0.300 m directly below it. We need to find the magnitude and sign of the unknown charge.
02

Apply Coulomb鈥檚 Law to find the unknown charge

Coulomb's Law states that the force between two charges is \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \), \( q_1 \) and \( q_2 \) are the charges in question, and \( r \) is the distance between them. Given that the force \( F = 0.200 \, \mathrm{N} \) and the distance \( r = 0.300 \, \mathrm{m} \), we start by listing the known values. We know \( q_1 = -0.550 \times 10^{-6} \, \mathrm{C} \), so we solve for \( q_2 \): \[0.200 = 8.99 \times 10^9 \frac{|-0.550 \times 10^{-6} q_2|}{(0.300)^2}\]Solve for \( q_2 \) to find the magnitude of the unknown charge.
03

Calculate the unknown charge

Rearrange the equation to solve for \( q_2 \):\[|-0.550 \times 10^{-6} q_2| = \frac{0.200 \times (0.300)^2}{8.99 \times 10^9} \approx 6.66 \times 10^{-12}\]Then solving for \( q_2 \):\[|q_2| = \frac{6.66 \times 10^{-12}}{0.550 \times 10^{-6}} \approx 1.21 \times 10^{-5} \, \mathrm{C} \]The charge is positive because the force is upward, indicating repulsion with the negative charge.
04

Confirm the sign and find the direction of forces

Since the force is upward and the charge below is unknown, the force鈥檚 direction is due to repulsive interaction from a positive charge. Thus, the unknown charge is positive \(+ 1.21 \times 10^{-5} \mathrm{C}\).
05

Apply Newton's Third Law to find the force on the known charge

Newton's third law states that every action has an equal and opposite reaction. Thus, the force that the unknown charge \(q_2\) exerts on \(-0.550 \mu C\) is equal in magnitude and opposite in direction to the initial force \(0.200 \, \mathrm{N}\). So, the magnitude is also \(0.200 \, \mathrm{N}\) but directed downward.

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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. It comes in two types: positive and negative.
  • Electrons carry a negative charge, while protons carry a positive charge.
  • Like charges repel each other, while opposite charges attract.
In the given problem, we have one known negative charge of electricity: \(-0.550 \mu C\). This charge exerts a force on an unknown charge placed 0.300 m directly below it. The goal is to determine the nature of the unknown charge. Since the known charge exerts an upward force, the unknown charge must be positive.Two like charges will repel, causing the force to act upward. Understanding electric charge helps in realizing that chemical reactions, electromagnetic phenomena, and even static cling are rooted in these basic interactions between charges. Electric charge serves as the basis for much of the forces we observe in physics, showing how fundamental it is to the world around us.
Force Interaction
When we discuss force interaction, we're talking about how two charges exert forces on each other due to their electric fields. Coulomb's Law quantifies this interaction. It states that the force between two charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them: \[F = k \frac{|q_1 q_2|}{r^2}\], where \(k\) is the Coulomb鈥檚 constant.
  • It allows us to calculate the force between two point charges.
  • The direction of force aligns with the principles of charge interaction: attraction or repulsion.
In our problem, the negative charge exerts 0.200 N force on the positive charge below it. Calculating Force Magnitude: Rearranging Coulomb's Law helps us find an unknown variable. Here, with a given force and distance, information on one charge allows calculation of the other. This approach is pivotal in solving scientific and engineering problems involving static electricity. Force interaction principles not only apply to small scales but are also essential in understanding more complex systems, such as electric circuits or electrostatic applications.
Newton's Third Law
Newton's Third Law of Motion is a cornerstone of classical mechanics. It states that for every action, there is an equal and opposite reaction. In the context of our exercise, this law is applicable in the analysis of the force exchanging between the charges.
  • When one charge exerts a force on another, the second charge exerts an equal and opposite force back on the first.
  • This means if one charge is pushing up, the other is pushing down with the same force magnitude.
For the exercise at hand, the 0.200 N force that the negative charge exerts upward indicates that the positive charge also exerts a downward 0.200 N force鈥攅qual in magnitude, but opposite in direction. Equal and Opposite Forces: Understanding Newton鈥檚 Third Law helps clarify why forces occur in pairs. If you're pushing against a wall, the wall pushes back on you with an equal force. In the world of electric charges, these principles ensure the stability and predictability of reactions. Comprehending this central law is crucial for fields ranging from mechanical engineering to physics, as it applies to all interactions regardless of their scale or complexity.

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Most popular questions from this chapter

\(\bullet\) A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter \(12.0 \mathrm{cm},\) giving it a charge of \(-15.0 \mu \mathrm{C}\) . Find the electric field (a) just inside the paint layer, (b) just outside the paint layer, and (c) 5.00 \(\mathrm{cm}\) out- side the surface of the paint laver.

\(\bullet$$\bullet\) Electrophoresis. Electrophoresis is a process used by biologists to separate dif- ferent biological molecules (such as pro- teins) from each other according to their ratio of charge to size. The materials to be separated are in a viscous solution that produces a drag force \(F_{\mathrm{D}}\) propor- tional to the size and speed of the molecule. We can express this relationship as \(F_{11}=K R v,\) where \(R\) is the radius of the molecule (modeled as being spherical), \(v\) is its speed, and \(K\) is a constant that depends on the viscosity of the solution. The solution is placed in an external electric field \(E\) so that the electric force on a particle of charge \(q\) is \(F=q E .\) (a) Show that when the electric field is adjusted so that the two forces (electrical and vis- cous drag) just balance, the ratio of \(q\) to \(R\) is \(K v / E\) . (b) Show that if we leave the electric field on for a time \(T\) , the distance \(x\) that the molecule moves during that time is \(x=(E T / k)(q / R)\) . (c) Sup- pose you have a sample containing three different biological mole- cules for which the molecular ratio \(q / R\) for material 2 is twice that of material 1 and the ratio for material 3 is three times that of mate- rial 1. Show that the distances migrated by these molecules after the same amount of time are \(x_{2}=2 x_{1}\) and \(x_{3}=3 x_{1} .\) In other words, material 2 travels twice as far as material \(1,\) and material 3 travels three times as far as material \(1 .\) Therefore, we have sepa- rated these molecules according to their ratio of charge to size. In practice, this process can be carried out in a special gel or paper, along which the biological molecules migrate. (See Figure 17.60 .) The process can be rather slow, requiring several hours for separa- tions of just a centimeter or so.

\(\bullet$$\bullet\) A point charge of \(-4.00 \mathrm{nC}\) is at the origin, and a second point charge of \(+6.00 \mathrm{nC}\) is on the \(x\) axis at \(x=0.800 \mathrm{m}\) . Find the magnitude and direction of the electric field at each of the following points on the \(x\) axis: (a) \(x=20.0 \mathrm{cm}\) , (b) \(x=1.20 \mathrm{m},(\mathrm{c}) x=-20.0 \mathrm{cm} .\)

\(\bullet\) A particle has a charge of \(-3.00 \mathrm{nC}\) . (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 m directly above it. (b) At what distance from the parti- cle does its electric field have a magnitude of 12.0 \(\mathrm{N} / \mathrm{C}\) ?

\(\bullet\) Signal propagation in neurons. Neurons are components of the nervous system of the body that transmit signals as elec- trical impulses travel along their length. These impulses propa- gate when charge suddenly rushes into and then out of a part of the neutron called an axon. Measurements have shown that, during the inflow part of this cycle, approximately \(5.6 \times 10^{11} \mathrm{Na}^{+}\) (sodium ions) per meter, each with charge \(+e\) enter the axon. How many coulombs of charge enter a 1.5 \(\mathrm{cm}\) length of the axon during this process?
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