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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 \( Q \) is approximately \( 1.112 \times 10^{-9} \text{ C} \).

Step by step solution

01

Identify the Necessary Forces

To have circular motion, the particle must experience a centripetal force. This force can come from the electric force exerted by the charge at the origin. We need to ensure that the electric force provides the needed centripetal force.
02

Express Centripetal Force Requirement

The centripetal force required for circular motion is given by the formula \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the particle, \( v \) is its velocity, and \( r \) is the radius of the circular path. Here, \( v = 50.0 \text{ m/s} \) and \( r = 0.2 \text{ m} \), and \( m = 0.8 \text{ g} = 0.0008 \text{ kg} \). Thus, \( F_c = \frac{0.0008 \times (50)^2}{0.2} \).
03

Calculate Centripetal Force

Substitute values: \( F_c = \frac{0.0008 \times 2500}{0.2} = \frac{2}{0.2} = 10 \text{ N} \). So the particle requires a force of 10 N to maintain circular motion.
04

Relate Electric Force to Charge

The electric force between two charges is given by Coulomb's law: \( F_e = \frac{k|Qq|}{r^2} \), where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is Coulomb's constant, \( Q \) is the charge at the origin we need to find, \( q = 4.00 \mu C = 4.00 \times 10^{-6} \text{ C} \), and \( r = 0.2 \text{ m} \). Set \( F_e \) equal to \( F_c \).
05

Solve for Charge Q

Set the electric force equal to the centripetal force: \[ \frac{k|Qq|}{r^2} = 10 \]. Substitute known values: \[ \frac{8.99 \times 10^9 \times |Q| \times 4.00 \times 10^{-6}}{(0.2)^2} = 10 \]. Simplify and solve for \( |Q| \): \[ |Q| = \frac{10 \times (0.2)^2}{8.99 \times 10^9 \times 4.00 \times 10^{-6}} \].
06

Perform Mathematical Calculations

Calculate \( |Q| \) using the equation: \[ |Q| = \frac{0.04}{35.96 \times 10^{-3}} = \frac{0.04}{0.03596} \approx 1.112 \times 10^{-9} \text{ C} \].

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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 the force required to keep an object moving in a circular path. This force always points towards the center of the circle and keeps a moving object on its circular path.
The formula for centripetal force is given by:
  • \( F_c = \frac{mv^2}{r} \)
This formula shows that the force depends on:
  • Mass \( m \) of the object.
  • Velocity \( v \) of the object.
  • Radius \( r \) of the circle.
In our problem, the particle requires a centripetal force provided not by a rope or gravity, but by an electric force from another charge. This is a unique way to achieve circular motion through electric interactions.
We find the required centripetal force using the given mass, speed, and radius of the path.
Coulomb's Law
Coulomb's Law describes the electric force between two changes. It plays a key role in determining how forces between charged particles interact.
The law states that the electric force \( F_e \) between two charges \( Q \) and \( q \) is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance \( r \) between them:
  • \( F_e = \frac{k|Qq|}{r^2} \)
Here, \( k \) is Coulomb's constant \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
In the context of our problem, we use Coulomb's Law to establish the electric force that has to match the centripetal force required to keep the particle in circular motion.
Electric Force
Electric force is a fundamental interaction between charged particles. It can either bring particles together or push them apart, depending on the kind of charges involved.
The force is powerful and can be calculated using Coulomb's Law, as shown earlier. It's determined by:
  • Magnitude and sign of both charges.
  • Distance separating the charges.
  • The constant \( k \), which defines the strength of the electric interaction in a vacuum.
In the given exercise, electric force acts as the necessary centripetal force. This balance between forces allows the particle to maintain its path around the fixed charge successfully. Understanding electric forces thus provides insight into the invisible pushes and pulls that shape particle dynamics in fields like both classical and modern physics.

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Most popular questions from this chapter

The charges and coordinates of two charged particles held fixed in an \(x y\) plane are \(q_{1}=+3.0 \mu \mathrm{C}, x_{1}=3.5 \mathrm{~cm}, y_{1}=0.50 \mathrm{~cm}\), and \(q_{2}=-4.0 \mu \mathrm{C}, x_{2}=-2.0 \mathrm{~cm}, y_{2}=1.5 \mathrm{~cm}\). Find the (a) magni- tude and (b) direction of the electrostatic force on particle 2 due to particle 1 . At what (c) \(x\) and (d) \(y\) coordinates should a third particle of charge \(q_{3}=+4.0 \mu \mathrm{C}\) be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

An electron is in a vacuum near Earth's surface and located at \(y=0\) on a vertical \(y\) axis. At what value of \(y\) should a second electron be placed such that its electrostatic force on the first electron balances the gravitational force on the first electron?

What must be the distance between point charge \(q_{1}=\) \(26.0 \mu \mathrm{C}\) and point charge \(q_{2}=-47.0 \mu \mathrm{C}\) for the electrostatic force between them to have a magnitude of \(5.70 \mathrm{~N} ?\)

A current of \(0.300\) A through your chest can send your heart into fibrillation, ruining the normal rhythm of heartbeat and disrupting the flow of blood (and thus oxygen) to your brain. If that current persists for \(2.00 \mathrm{~min}\), how many conduction electrons pass through your chest?

In crystals of the salt cesium chloride, cesium ions \(\mathrm{Cs}^{+}\) form the eight corners of a cube and a chlorine ion \(\mathrm{Cl}^{-}\) is at the cube's center (Fig. 21-36). The edge length of the cube is \(0.40 \mathrm{~nm}\). The \(\mathrm{Cs}^{+}\) ions are each deficient by one electron (and thus each has a charge of \(+e\) ), and the \(\mathrm{Cl}^{-}\) ion has one excess electron (and thus has a charge of \(-e\) ). (a) What is the magnitude of the net electrostatic force exerted on the \(\mathrm{Cl}^{-}\) ion by the eight \(\mathrm{Cs}^{+}\) ions at the corners of the cube? (b) If one of the \(\mathrm{Cs}^{+}\) ions is missing, the crystal is said to have a defect; what is the magnitude of the net electrostatic force exerted on the \(\mathrm{Cl}^{-}\) ion by the seven remaining \(\mathrm{Cs}^{+}\) ions?
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