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Two \(1.0 \mathrm{kg}\) masses are \(1.0 \mathrm{m}\) apart on a frictionless table. Each has \(+1.0 \mu C\) of charge. a. What is the magnitude of the electric force on one of the masses? b. What is the initial acceleration of each mass if they are released and allowed to move?

Short Answer

Expert verified
a. The magnitude of the electric force on each mass is \(8.99 N\). b. The initial acceleration of each mass is \(8.99 m/s^2\).

Step by step solution

01

Calculate the magnitude of the electric force

Using the Coulomb's Law which gives the force between two charges as \(F = k \frac{|q1q2|}{r^2}\), where \(k = 8.99x10^9 Nm^2/c^2\) is Coulomb's constant, \(q1 = q2 = 1.0 \mu C = 1.0x10^-6 C\) are the charges and \(r = 1.0 m\) is the distance between them. Therefore, substituting these values into the equation gives \[F = 8.99x10^9 \frac{ (1.0x10^-6)^2}{(1.0)^2} = 8.99 N\].
02

Calculate the initial acceleration

Next, apply Newton's second law \(F = ma\) where \(F = 8.99 N\) is the force calculated from step 1 and \(m = 1.0 kg\) is the mass of the object. Therefore, rearranging the equation for acceleration gives \(a = \frac{F}{m}\) Hence, substituting these values into the equation gives \[a = \frac{8.99}{1.0} = 8.99 m/s^2\]. So the initial acceleration of each mass is \(8.99 m/s^2\).

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Key Concepts

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

Coulomb's Law
Let's dive into the heart of electric forces: Coulomb's Law. It's a crucial concept in physics for calculating the force between two charged objects. The formula is expressed as:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
Here, \(k\) stands for Coulomb's constant, a fundamental constant in nature valued at \(8.99 \times 10^9 \ \mathrm{N\cdot m^2/C^2}\). The values \(q_1\) and \(q_2\) represent the magnitudes of the two charges involved, and \(r\) is the distance separating them. The equation shows that the electric force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Addition of importance, the positive charges in our exercise repel each other, yielding an electric force because like charges push away from one another. Grasping this formula and its components helps you solve problems involving any charged objects placed in proximity.
Newton's Second Law
Newton's Second Law links force, mass, and acceleration, giving us deeper insights into motion. The formula is succinctly captured as:
  • \( F = ma \)
In this context, \(F\) is the net force acting on an object measured in newtons (N), \(m\) is the object's mass measured in kilograms (kg), and \(a\) is the acceleration, indicated in meters per second squared (m/s²).
Understanding this law offers clarity on how force affects motion. In our exercise, after computing the electric force using Coulomb's Law, we pivot towards understanding how this force acts on the object. By rearranging the equation to \(a = \frac{F}{m}\), we can determine the object's acceleration, revealing how quickly it will start moving due to the electric force.
This law is not only instrumental in tackling physics problems but is foundational to understanding dynamics in many real-world scenarios.
Charge Interaction
Charge interaction forms the backbone of understanding electric forces. Charges are a property of matter that denote either positive or negative electrical units.
Two principal types of interactions exist:
  • Like charges (positive-positive or negative-negative) repel each other.
  • Unlike charges (positive-negative) attract each other.
In terms of our problem, the two identical charges, both being positive \((+1.0 \ \mu C)\), repel, generating an electric force. This repulsion causes the separation between the two, as they exert forces on each other to push away.
These interactions between charges are governed by fundamental physics rules, making them predictable and calculable. By acknowledging the types and nature of these interactions, one gains insights into predicting the behavior of charged objects.
Acceleration Calculation
Taking on the task of calculating acceleration means understanding the changes in an object's motion driven by force. From the previous steps, we've identified both the force and the mass.
  • Using the rearranged Newton's Second Law, \(a = \frac{F}{m}\), we substitute the values to find acceleration.
Given our problem, with a known force of \(8.99\ N\) and mass \(1.0\ kg\), the calculation becomes:
  • \(a = \frac{8.99}{1.0} = 8.99\, \mathrm{m/s^2}\).
This shows how rapidly the masses will begin to accelerate once released, highlighting the potential of electric forces in motion. Such calculations are pivotal in physics for predicting motion, allowing us to anticipate how quickly velocities change when subjected to different forces.

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Most popular questions from this chapter

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two \(1.8 \mathrm{mg}\) drops each have a charge of \(+25 \mathrm{pC} ;\) these are typical values. The centers of the droplets are at the same height and \(0.40 \mathrm{cm}\) apart. What is the approximate electric force between them? What horizontal acceleration does this force produce on the droplets?

A plastic rod that has been charged to \(-15.0 \mathrm{nC}\) touches a metal sphere. Afterward, the rod's charge is \(-10.0 \mathrm{nC}\). a. What kind of charged particle was transferred between the rod and the sphere, and in which direction? That is, did it move from the rod to the sphere or from the sphere to the rod?

I Two protons are \(2.0 \mathrm{fm}\) apart. \((1 \mathrm{fm}=1\) femtometer \(=\) \(\left.1 \times 10^{-15} \mathrm{m} .\right)\) a. What is the magnitude of the electric force on one proton due to the other proton? b. What is the magnitude of the gravitational force on one proton due to the other proton? c. What is the ratio of the electric force to the gravitational force?

A platypus foraging for prey can detect an electric field as small as \(0.002 \mathrm{N} / \mathrm{C}\). To give an idea of the sensitivity of the platypus's electric sense, how far from a \(+10 \mathrm{nC}\) point charge does the field have this magnitude?

A plastic rod is charged to \(-20 \mathrm{nC}\) by rubbing. a. Have electrons been added to the rod or protons removed? Explain. b. How many electrons have been added or protons removed?

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