Problem 45 - University Physics with Modern Physics

Question

\(A+2.00-n C\) 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: \((\text { i) } x=0.200 \mathrm{m} ; \text { (ii) } x=1.20 \mathrm{m} ; \text { (iii) } x=\) \(-0.200 \mathrm{m} .\) (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 given points are: (i) strong and towards the negative charge, (ii) towards the negative charge, (iii) towards the positive charge. Force on the electron follows the field direction.

Step 1: Understand the Problem

We need to find the electric field, denoted as \( E \), at three specified points along the x-axis, due to two point charges: one \( +2.00\, \mathrm{nC} \) at the origin \((0,0)\) and one \( -5.00\, \mathrm{nC} \) at \(x=0.800\, \mathrm{m}\). Then, we need to find the force on an electron at those points. Use the formula for the electric field produced by a point charge: \[ E = \frac{k |q|}{r^2} \]where \( k \) is Coulomb's constant \((8.99\times10^9 \mathrm{Nm}^2/\mathrm{C}^2)\), \(q\) is the charge, and \(r\) is the distance from the charge.

Step 2: Electric Field at \( x = 0.200 \mathrm{m} \)

Calculate the electric field at \( x = 0.200 \mathrm{m} \):1. Distance from \( +2.00\, \mathrm{nC} \) charge at the origin: \( r_1 = 0.200\, \mathrm{m} \)2. Distance from \( -5.00\, \mathrm{nC} \) charge: \( r_2 = 0.800 - 0.200 = 0.600\, \mathrm{m} \)3. Electric field from \( +2.00\, \mathrm{nC} \) :\[ E_1 = \frac{(8.99 \times 10^9)(2.00 \times 10^{-9})}{(0.200)^2} \]4. Electric field from \( -5.00\, \mathrm{nC} \) :\[ E_2 = \frac{(8.99 \times 10^9)(5.00 \times 10^{-9})}{(0.600)^2} \] (points in the opposite direction)5. Net Electric Field at \( x = 0.200 \mathrm{m} \):\[ E_{net} = E_1 - E_2 \]Insert values to compute \( E_{net} \).

Step 3: Electric Field at \( x = 1.20 \mathrm{m} \)

For \( x = 1.20 \mathrm{m} \):1. Distance from \( +2.00\, \mathrm{nC} \) charge: \( r_1 = 1.20 \mathrm{m} \)2. Distance from \( -5.00\, \mathrm{nC} \) charge: \( r_2 = 1.20 - 0.800 = 0.400 \mathrm{m} \)3. Electric field from \( +2.00\, \mathrm{nC} \) :\[ E_1 = \frac{(8.99 \times 10^9)(2.00 \times 10^{-9})}{(1.20)^2} \]4. Electric field from \( -5.00\, \mathrm{nC} \):\[ E_2 = \frac{(8.99 \times 10^9)(5.00 \times 10^{-9})}{(0.400)^2} \]5. Net Electric Field at \( x = 1.20 \mathrm{m} \):\[ E_{net} = E_2 - E_1 \]Insert values to compute \( E_{net} \).

Step 4: Electric Field at \( x = -0.200 \mathrm{m} \)

At \( x = -0.200 \mathrm{m} \):1. Distance from \( +2.00\, \mathrm{nC} \) charge: \( r_1 = 0.200 \mathrm{m} \)2. Distance from \( -5.00\, \mathrm{nC} \) charge: \( r_2 = 0.800 + 0.200 = 1.00 \mathrm{m} \)3. Electric field from \( +2.00\, \mathrm{nC} \):\[ E_1 = \frac{(8.99 \times 10^9)(2.00 \times 10^{-9})}{(0.200)^2} \]4. Electric field from \( -5.00\, \mathrm{nC} \):\[ E_2 = \frac{(8.99 \times 10^9)(5.00 \times 10^{-9})}{(1.00)^2} \] (opposite direction)5. Net Electric Field at \( x = -0.200 \mathrm{m} \):\[ E_{net} = E_1 - E_2 \]Calculate the net electric field.

Step 5: Calculate Force on Electron

For an electron at each position, the force \( F \) is related to the net electric field by the equation \( F = e \cdot E_{net} \), where \( e = 1.60 \times 10^{-19} \mathrm{C} \).1. Compute the force at \( x = 0.200 \mathrm{m} \) using the corresponding \( E_{net} \) value.2. Compute the force at \( x = 1.20 \mathrm{m} \). 3. Compute the force at \( x = -0.200 \mathrm{m} \).Substitute the respective \( E_{net} \) to find the force at each position.

Key Concepts

Point Charge

When we talk about a point charge, we are describing a charged object that is so small that its size doesn't affect calculations relating to electric fields or forces. Instead of worrying about the size or shape of the charged object, we only care about its position and the amount of charge it carries. This simplification helps us focus on understanding how electric fields work and interact with other charges.

In physics, when dealing with point charges, we denote the charge with the letter \( q \). The charge could be positive or negative, influencing whether the electric field around it repels or attracts other charges. For example, a positive point charge emits an electric field that pushes other positive charges away, whereas it attracts negative charges. In this exercise, we had two point charges: a \( +2.00 \, \mathrm{nC} \) charge at the origin of the axis and a \( -5.00 \, \mathrm{nC} \) charge located further down the line at \( x = 0.800 \, \mathrm{m} \). Each of these charges affects the electric field strength at various points along the x-axis, which we calculated in subsequent steps.

Coulomb's Law

Coulomb's Law describes how charged objects interact with each other, defining the force between two point charges. The law is expressed mathematically as: \[ F = \frac{k |q_1 q_2|}{r^2} \]where \( F \) is the force between the charges, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.

Coulomb's Law allows us to understand the nature of attraction or repulsion between charges. A positive force indicates repulsion (like charges), while a negative force suggests attraction (opposite charges). Applied to electric fields, this law helps us calculate the electric field strength \( E \) that a point charge creates at any location in space using the expression: \[ E = \frac{k |q|}{r^2} \]. In this exercise, the charges involved created electric fields that we calculated with Coulomb's Law at specific positions, which later helped determine the electric force on an electron.

Electric Force on Electron

The electric force on an electron is a consequence of an electric field acting upon it. Since an electron has a negative charge, it moves towards positive fields and away from negative ones. The force on a charge in an electric field is calculated using the equation:\[ F = e \cdot E_{net} \]where \( e \) is the charge of the electron (\( -1.60 \times 10^{-19} \, \mathrm{C} \)) and \( E_{net} \) is the net electric field at a given point.

In our exercise, we determined the net electric field \( E_{net} \) at different positions on the x-axis due to the presence of the two point charges. We then used these values to calculate the force acting on an electron placed at each position. This exercise illustrates the practical application of electric fields, where forces on charged particles like electrons can be predicted and calculated.