Problem 45 - University Physics with Modern Physics

Question

\(A+2.00-n C \quad\) 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).

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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.

Step 5: 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).

Key Concepts

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.