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Chapter 21: Problem 52

A particle of charge \(Q\) is fixed at the origin of an \(x y\) coordinate system. At \(t=0\) a particle \((m=0.800 \mathrm{~g}, q=4.00 \mu \mathrm{C})\) is located on the \(x\) axis at \(x=20.0 \mathrm{~cm}\), moving with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. For what value of \(Q\) will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

Short Answer

Expert verified
The value of \( Q \) needed is approximately \( 5.56 \times 10^{-8} \, C \).

Step by step solution

01

Identify the Forces

The moving particle will experience an electrostatic force due to the charge \( Q \) at the origin. The force is given by Coulomb's law: \( F = \frac{k \, |Q \, q|}{r^2} \), where \( k \) is Coulomb's constant (\(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)), \( q = 4.00 \, \mu\text{C} \), and \( r = 0.2 \, \text{m} \).
02

Analyze Circular Motion Conditions

For the particle to undergo circular motion, the centripetal force \( F_c \) must equal the electrostatic force \( F \). The centripetal force required is given by \( F_c = \frac{m v^2}{r} \), where \( m = 0.800 \, \text{g} = 0.0008 \, \text{kg} \) and \( v = 50.0 \, \text{m/s} \).
03

Set up the Equation

Equate the centripetal force to the electrostatic force: \[ \frac{m v^2}{r} = \frac{k \, |Q \, q|}{r^2} \].
04

Solve for Q

Simplify and rearrange the equation to find \( Q \): \[ |Q| = \frac{m v^2 r}{k \, |q|} \]. Substitute the known values: \( m = 0.0008 \, \text{kg} \), \( v = 50.0 \, \text{m/s} \), \( r = 0.2 \, \text{m} \), \( q = 4.00 \times 10^{-6} \, \text{C} \), \( k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
05

Calculate the Value of Q

Substitute into the formula: \[ |Q| = \frac{0.0008 \, (50)^2 \, 0.2}{8.99 \times 10^9 \, (4 \times 10^{-6})} \]. Simplify to find \( |Q| \approx 5.56 \times 10^{-8} \, \text{C} \).

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Key Concepts

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

Centripetal Force
Centripetal Force is an essential concept when studying circular motion, as it is the force that keeps an object moving in a circle. This force constantly pulls the object toward the center of the circular path it is following.
The way centripetal force works is quite fascinating:
  • This force acts perpendicular to the velocity of the object.
  • The formula for centripetal force is given by \( F_c = \frac{m v^2}{r} \), where \( m\) is the mass, \( v\) is the velocity, and \( r\) is the radius of the circle.
  • Importantly, without this force, the object would fly off in a straight line due to inertia.
In the context of the exercise, centripetal force ensures that the particle keeps a circular path around the point where charge \( Q \) is located. This aligns with the electrostatic attraction exerted by the particle at the origin.
Coulomb's Law
Coulomb's Law is fundamental in understanding how charges interact. It describes the electrostatic force between two point charges. The force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Here's a breakdown of the key elements:
  • The formula is expressed as \( F = \frac{k \, |Q \, q|}{r^2} \), where \( k \) is Coulomb's constant, \( Q\) and \( q\) are the charges, and \( r\) is the distance.
  • This law helps calculate the magnitude of electrostatic forces in various scenarios.
  • The force can be attractive or repulsive, depending on the signs of the charges involved.
In the given exercise, Coulomb's law is crucial as it quantifies the electrostatic force that provides the necessary centripetal force to maintain the particle's circular motion around the origin.
Circular Motion
When a particle moves in a circular path, it experiences an acceleration directed towards the center of the circle. This type of movement is known as circular motion, which is prevalent in both natural forces and mechanical systems.
Here are some intriguing aspects of circular motion:
  • In circular motion, an object travels at a constant speed along a circular path.
  • However, even though the speed remains constant, the velocity changes due to its continuous change in direction.
  • This change in direction results in centripetal acceleration, necessary for sustaining circular motion.
In the problem discussed, we deal with circular motion driven by electrostatic forces. The attractive force between the particle with charge \( Q \) at the origin and the moving charge ensures that the particle stays on its curved path, conforming to the laws of circular motion.

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Most popular questions from this chapter

A charge of \(6.0 \mu \mathrm{C}\) is to be split into two parts that are then separated by \(3.0 \mathrm{~mm}\). What is the maximum possible magnitude of the electrostatic force between those two parts?

Four particles form a square. The charges are \(q_{1}=+Q, q_{2}=q_{3}=q\), and \(q_{4}=-2.00 Q .\) What is \(q / Q\) if the net electrostatic force on particle 1 is zero?

Shows four identical conducting spheres that are actually well separated from one another. Sphere \(W\) (with an initial charge of zero) is touched to sphere \(A\) and then they are separated. Next, sphere \(W\) is touched to sphere \(B\) (with an initial charge of \(-32 e\) ) and then they are separated. Finally, sphere \(W\) is touched to sphere \(C\) (with an initial charge of \(+48 e\) ), and then they are separated. The final charge on sphere \(W\) is \(+18 e\). What was the initial charge on sphere \(A\) ?

We know that the negative charge on the electron and the positive charge on the proton are equal. Suppose, however, that these magnitudes differ from each other by \(0.00010 \%\). With what force would two copper coins, placed \(1.0 \mathrm{~m}\) apart, repel each other? Assume that each coin contains \(3 \times 10^{22}\) copper atoms. (Hint: \(\mathrm{A}\) neutral copper atom contains 29 protons and 29 electrons.) What do you conclude?

Two tiny conducting balls of identical mass \(m\) and identical charge \(q\) hang from nonconducting threads of length \(L\). Assume that \(\theta\) is so small that \(\tan \theta\) can be replaced by its approximate equal, \(\sin \theta\). (a) Show that $$ x=\left(\frac{q^{2} L}{2 \pi \varepsilon_{0} m g}\right)^{1 / 3} $$ gives the equilibrium separation \(x\) of the balls. (b) If \(L=120 \mathrm{~cm}, m=10 \mathrm{~g}\), and \(x=5.0 \mathrm{~cm}\), what is \(|q| ?\)
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