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A \(+2.00-n C\) point charge is at the origin, and a second \(-5.00-n C\) point charge is on the \(x\)-axis at \(x=0.800 \mathrm{~m} .\) (a) Find the electric field (magnitude and direction) at each of the following points on the \(x\)-axis: (i) \(x=0.200 \mathrm{~m}\); (ii) \(x=-0.200 \mathrm{~m}\). (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

Short Answer

Expert verified
(a) At \(x=0.200 \mathrm{~m}\), \(E = 324.64 \text{ N/C (right)}\); at \(x=-0.200 \mathrm{~m}\), \(E = 404.55 \text{ N/C (left)}\). (b) Forces are \(5.19 \times 10^{-17} \text{ N (left)}\) and \(6.47 \times 10^{-17} \text{ N (right)}\).

Step by step solution

01

Identify Given Charges and Positions

The positive charge is \(+2.00 \text{ nC}\) located at the origin \((x=0)\) and the negative charge is \(-5.00 \text{ nC}\) at \(x=0.800 \text{ m}\). Points to evaluate the electric field: (i) \(x=0.200 \text{ m}\), (ii) \(x=-0.200 \text{ m}\).
02

Recall the Formula for Electric Field

The electric field \(E\) due to a point charge \(q\) at a distance \(r\) is given by: \[ E = \frac{k |q|}{r^2} \]where \(k = 8.99 \times 10^9 \text{ N}\cdot \text{m}^2/\text{C}^2\) is the Coulomb's constant.
03

Calculate Electric Field at x = 0.200 m

For the positive charge: distance \(r_1 = 0.200 \text{ m}\), so \[ E_1 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-9}}{(0.200)^2} = 449.5 \text{ N/C}\text{ (rightward)} \]For the negative charge: distance \(r_2 = 0.600 \text{ m}\), so \[ E_2 = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-9}}{(0.600)^2} = 124.86 \text{ N/C}\text{ (leftward)} \]Net electric field at \(x=0.200 \text{ m}\) is \[ E = E_1 - E_2 = 449.5 - 124.86 = 324.64 \text{ N/C (rightward)} \]
04

Calculate Electric Field at x = -0.200 m

For the positive charge: distance \(r_3 = 0.200 \text{ m}\), so \[ E_3 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-9}}{(0.200)^2} = 449.5 \text{ N/C (leftward)} \]For the negative charge: distance \(r_4 = 1.000 \text{ m}\), so \[ E_4 = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-9}}{(1.000)^2} = 44.95 \text{ N/C (rightward)} \]Net electric field at \(x=-0.200 \text{ m}\) is \[ E = E_3 - E_4 = 449.5 - 44.95 = 404.55 \text{ N/C (leftward)} \]
05

Calculate Electric Force on Electron at x=0.200 m

Using \(F = e \cdot E\) where \(e = 1.60 \times 10^{-19} \text{ C}\), for \(x=0.200 \text{ m}\) \[ F = 1.60 \times 10^{-19} \times 324.64 = 5.19 \times 10^{-17} \text{ N (leftward)} \]
06

Calculate Electric Force on Electron at x=-0.200 m

Using \(F = e \cdot E\), for \(x=-0.200 \text{ m}\) \[ F = 1.60 \times 10^{-19} \times 404.55 = 6.47 \times 10^{-17} \text{ N (rightward)} \]

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle that explains how electric charges interact with each other. This law states that the electric force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. In simple terms:
  • The greater the charges, the stronger the force.
  • The greater the distance, the weaker the force.
The formula to express Coulomb's law is given by:\[ F = \frac{k |q_1 q_2|}{r^2} \]where:
  • \( F \) is the magnitude of force between the charges,
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2 \),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the two charges,
  • \( r \) is the distance between the charges.
This law helps us understand the electric interactions, like those happening in atoms and molecules. It is the groundwork for calculating forces, fields, and potentials.
Electric Force on Electron
An electron, being a negatively charged particle, experiences a force when placed in an electric field. This force is known as the electric force and is crucial in determining how charged particles move within an electric field. The expression for calculating the force on an electron is:\[ F = e \cdot E \] where:
  • \( F \) is the electric force on the electron,
  • \( e \) is the charge of the electron \( (1.60 \times 10^{-19} \text{ C}) \),
  • \( E \) is the electric field strength at that point.
The direction of the force on an electron will be opposite the direction of the electric field because the electron carries a negative charge. This concept is foundational in fields like electronics and particle physics, where moving charges are in complex fields.
Electric Field Calculation
The electric field is a vector field that describes the force per unit charge exerted on any other charge placed within the field. To calculate the electric field due to a point charge, we use:\[ E = \frac{k |q|}{r^2} \]where:
  • \( E \) is the electric field magnitude,
  • \( k \) is the Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2 \).
  • \( q \) is the point charge producing the field,
  • \( r \) is the distance from the charge to the point of interest.
The direction of the electric field is radially outwards for positive charges and inwards for negative charges. This calculation is vital in understanding the influence of charges on each other, defining the space around a charge where its effect can be felt - kind of like an invisible map guiding another charge's behavior.
Superposition Principle in Electric Fields
The superposition principle states that the net electric field at a given point due to multiple charges is the vector sum of the electric fields produced by each charge independently. In mathematical terms, if you have multiple charges, the total electric field \( E_{net} \) can be calculated as:\[ E_{net} = E_1 + E_2 + E_3 + ext{...} + E_n \]where each \( E_n \) represents the electric field due to an individual charge at a particular location. When applying this principle, make sure to consider:
  • Direction: Electric fields have direction. Use vector addition for accurate results.
  • Magnitude: Calculate the strength of each field component accurately using the distance from each charge to the observation point.
This principle allows for understanding complex arrangements of charges by simplifying them into singular interactions. It's like solving a puzzle using smaller, easier parts and is applied in everything from calculating fields around molecular structures to designing circuits.

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Most popular questions from this chapter

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(2.25 \mathrm{~cm}\) distant from the first, in a time interval of \(1.50 \times 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

A tiny electric dipole \(\left(\mathrm{H}_{2} \mathrm{O}\right.\) modecule) of dipole moment \(\vec{p}\) is placed at a distance \(r\) form an infinitely long wire, with its \(\vec{p}\) normal to the wire in the same plane. If the linear charge density of the wire is \(\lambda\), find the electrostatic force acting on the dipole is equal to

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L\). Find the magnitude and direction of the net force on a point charge \(-3 q\) placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3 q\) charge by each of the other three charges.

At \(t=0\) a very small object with mass \(0.400 \mathrm{mg}\) and charge \(+9.00 \mu \mathrm{C}\) is traveling at \(125 \mathrm{~m} / \mathrm{s}\) in the \(-x\)-direction. The charge is moving in a uniform electric field that is in the \(+y\)-direction and that has magnitude \(E=895 \mathrm{~N} / \mathrm{C}\). The gravitational force on the particle can be neglected. How far is the particle from the origin at \(t=7.00 \mathrm{~ms} ?\)

A thin disk with a circular hole at its center, called an annulus, has inner radius \(R_{1}\) and outer radius \(R_{2}\) (Fig. P1.57). The disk has a uniform positive surface charge density \(\sigma\) on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the \(y z\)-plane, with its center at the origin. For an arbitrary point on the \(x\)-axis (the axis of the annulus), find the magnitude and direction of the electric field \(\overrightarrow{\boldsymbol{E}}\). Consider points both above and below the annulus in Fig. P1.56. (c) Show that at points on the \(x\)-axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is "sufficiently close"? (d) A point particle with mass \(m\) and negative charge \(-q\) is free to move along the \(x\)-axis (but cannot move off the axis). The particle is originally placed at rest at \(x=0.01 R_{1}\) and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.)

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