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Chapter 21: Problem 45

\(A+2.00-n C \quad\) point charge is at the origin, and a second \(-5.00\) -n \(\mathrm{C}\) 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) \(x=1.20 \mathrm{m} ;\) (iii) \(x=-0.200 \mathrm{m} .\) (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
The electric fields at (i) 324.64 N/C, right; (ii) 268.44 N/C, left; (iii) 494.45 N/C, right. Forces on electron are opposite directions to fields.

Step by step solution

01

Identify the Electric Field Formula

The electric field due to a point charge \(q\) at a distance \(r\) is given by Coulomb's Law: \[ E = \frac{k \cdot q}{r^2} \]where \(k = 8.99 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2\) is the Coulomb's constant.
02

Calculate Electric Field at x=0.200 m

Calculate the distance of \(+2.00 \, \mathrm{nC}\) charge from point \(x = 0.200 \, \mathrm{m}\), \(r_1 = 0.200 \, \mathrm{m}\).Calculate the distance of \(-5.00 \, \mathrm{nC}\) charge from point \(x = 0.200 \, \mathrm{m}\), \(r_2 = 0.600 \, \mathrm{m}\).Apply the formula: \[ E = \frac{k \cdot |q|}{r^2} \]Calculate \(E_1\) due to \(+2.00 \, \mathrm{nC}\): \[ E_1 = \frac{8.99 \times 10^9 \cdot 2.00 \times 10^{-9}}{(0.200)^2} = 449.5 \, \mathrm{N/C}\]Calculate \(E_2\) due to \(-5.00 \, \mathrm{nC}\): \[ E_2 = \frac{8.99 \times 10^9 \cdot 5.00 \times 10^{-9}}{(0.600)^2} = 124.86 \, \mathrm{N/C}\]Because \(+q\) creates an outward field and \(-q\) creates an inward field:The net electric field = \(E_1 - E_2 = 449.5 - 124.86 = 324.64 \, \mathrm{N/C}\), directed to the right.
03

Calculate Electric Field at x=1.20 m

Calculate the distance of \(+2.00 \, \mathrm{nC}\) charge from \(x = 1.20 \, \mathrm{m}\), \(r_1 = 1.20 \, \mathrm{m}\).Calculate the distance of \(-5.00 \, \mathrm{nC}\) charge from \(x = 1.20 \, \mathrm{m}\), \(r_2 = 0.400 \, \mathrm{m}\).Calculate \(E_1\) due to \(+2.00 \, \mathrm{nC}\): \[ E_1 = \frac{8.99 \times 10^9 \cdot 2.00 \times 10^{-9}}{(1.20)^2} = 12.49 \, \mathrm{N/C}\]Calculate \(E_2\) due to \(-5.00 \, \mathrm{nC}\): \[ E_2 = \frac{8.99 \times 10^9 \cdot 5.00 \times 10^{-9}}{(0.400)^2} = 280.93 \, \mathrm{N/C}\]The net electric field = \(E_2 - E_1 = 280.93 - 12.49 = 268.44 \, \mathrm{N/C}\), directed to the left.
04

Calculate Electric Field at x=-0.200 m

Calculate the distance of \(+2.00 \, \mathrm{nC}\) charge from \(x = -0.200 \, \mathrm{m}\), \(r_1 = 0.200 \, \mathrm{m}\).Calculate the distance of \(-5.00 \, \mathrm{nC}\) charge from \(x = -0.200 \, \mathrm{m}\), \(r_2 = 1.00 \, \mathrm{m}\).Calculate \(E_1\) due to \(+2.00 \, \mathrm{nC}\): \[ E_1 = \frac{8.99 \times 10^9 \cdot 2.00 \times 10^{-9}}{(0.200)^2} = 449.5 \, \mathrm{N/C}\]Calculate \(E_2\) due to \(-5.00 \, \mathrm{nC}\): \[ E_2 = \frac{8.99 \times 10^9 \cdot 5.00 \times 10^{-9}}{(1.00)^2} = 44.95 \, \mathrm{N/C}\]The net electric field = \(E_1 + E_2 = 449.5 + 44.95 = 494.45 \, \mathrm{N/C}\), directed to the right.
05

Calculate Electric Force on Electron

The force on a charge \(q\) in an electric field \(E\) is given by: \[ F = q_e \cdot E \]where \(q_e = -1.60 \times 10^{-19} \, \mathrm{C}\) is the charge of an electron. For point (i), \( F = -1.60 \times 10^{-19} \cdot 324.64 \, \mathrm{N/C} = -51.94 \times 10^{-19} \, \mathrm{N}\), direction is reversed (left).For point (ii), \( F = -1.60 \times 10^{-19} \cdot 268.44 \, \mathrm{N/C} = -42.95 \times 10^{-19} \, \mathrm{N}\), direction is reversed (right).For point (iii), \( F = -1.60 \times 10^{-19} \cdot 494.45 \, \mathrm{N/C} = -79.11 \times 10^{-19} \, \mathrm{N}\), direction is reversed (left).

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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 foundational concept in physics that describes the interaction between two charged objects. It 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. This can be expressed mathematically as: \[ F = k \frac{|q_1 \cdot q_2|}{r^2} \] where:
  • \( F \) is the magnitude of the electric force,
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \ \mathrm{Nm}^2/\mathrm{C}^2 \),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the point charges,
  • \( r \) is the distance between the centers of the two charges.
The force is attractive if the charges are of opposite signs and repulsive if they are of the same sign. Understanding Coulomb's Law is essential for solving problems involving electric fields and forces between charged particles.
Point Charge
A point charge is an idealized model of a charged particle considered as a point mass with an electric charge. It is often used in physics to simplify the study of electric fields and forces. Point charges are crucial in theoretical models because:
  • They allow for clear calculations and interpretations of electric field strength and direction.
  • Their electric field is radial and spherically symmetrical, emanating outward for positive charges and inward for negative charges.
In real-life applications, objects are not perfect point charges, but many can be approximated as such when they are small compared to the distances involved in the problem. In the given exercise, point charges are used to determine the electric field at specific points along the x-axis.
Electric Force
Electric force is a fundamental force that occurs between electrically charged particles. According to Coulomb's Law, this force can be attractive or repulsive depending on the signs of the charges involved. The force acting on a point charge in an electric field can be calculated with: \[ F = q \cdot E \] where:
  • \( F \) is the electric force,
  • \( q \) is the charge of the particle,
  • \( E \) is the electric field strength.
The direction of the electric force is dependent on the charge:
  • If the charge is positive, the force is in the direction of the electric field.
  • If the charge is negative, the force is in the opposite direction.
Understanding electric force is vital when examining how charged particles interact with electric fields, as these interactions are fundamental to the workings of electronic devices and many natural phenomena.
Coulomb's Constant
Coulomb's constant, denoted by \( k \), is a proportionality constant in Coulomb's Law that quantifies the strength of the electrostatic force between two point charges. Its value is approximately \( 8.99 \times 10^9 \ \mathrm{Nm}^2/\mathrm{C}^2 \). It plays a key role in calculating electric forces and fields because it determines the scale at which these interactions occur. Here are a few key points:
  • Coulomb's constant is derived from the permittivity of free space \( \varepsilon_0 \), as \( k = \frac{1}{4\pi\varepsilon_0} \).
  • Its large value indicates the significant strength of the force between charged particles compared to gravitational forces.
  • Coulomb's constant allows for the conversion of charge and distance into force, making it essential for practical calculations in electrostatics.
In essence, understanding Coulomb's constant is important for any study or application that involves electric charge interactions and allows for precise calculations in various electrodynamics problems.

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Most popular questions from this chapter

(a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A\) ? (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s},\) east. What is the speed of the proton at point \(B\) ?

Two very large horizontal sheets are 4.25 \(\mathrm{cm}\) apart and carry equal but opposite uniform surface charge densities of magnitude \(\sigma .\) You want to use these sheets to hold stationary in the region between them an oil droplet of mass 324\(\mu\) that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should \(\sigma\) be?

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=\) \(0.600 \mathrm{m}, y=0.800 \mathrm{m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

Two small spheres spaced 20.0 \(\mathrm{cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{N} ?\)

Particles in a Gold Ring. You have a pure (24 karat) gold ring with mass 17.7 \(\mathrm{g}\) . Gold has atomic mass of 197 \(\mathrm{g} / \mathrm{mol}\) and an atomic number of \(79 .\) (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?
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