Chapter 2: Problem 2
Point charges of \(1 \mathrm{nC}\) and \(-2 \mathrm{nC}\) are located at \((0,0,0)\) and \((1,1,1)\), respectively, in free space. Determine the vector force acting on each charge.
Short Answer
Expert verified
In summary, the vector force acting on the \(1 \mathrm{nC}\) charge is \(3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\mathrm{N}\) and the vector force acting on the \(-2 \mathrm{nC}\) charge is \(-3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\mathrm{N}\). These two forces are equal in magnitude but opposite in direction.
Step by step solution
01
Understand Coulomb's Law
Coulomb's law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it can be written as:
\(F = k\frac{|q_1q_2|}{r^2}\)
where \(F\) is the electrostatic force, \(k\) is Coulomb's constant (\(8.99 × 10^9 Nm^2/C^2\)), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between them.
02
Convert the charges to Coulombs
The charges given are in nanocoulombs (nC). We should first convert them to Coulombs (C) before proceeding:
\(1 \mathrm{nC} = 10^{-9} \mathrm{C}\)
So, \(q_1 = 1 \times 10^{-9} \mathrm{C}\) and \(q_2 = -2 \times 10^{-9} \mathrm{C}\).
03
Calculate the distance between the charges
Since the charges are located at \((0,0,0)\) and \((1,1,1)\), we can use the distance formula to find the distance between them:
\(r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} = \sqrt{(1-0)^2 + (1-0)^2 + (1-0)^2} = \sqrt{3}\)
04
Apply Coulomb's Law to find the magnitude of force
Using the values of \(q_1\), \(q_2\), and \(r\), we can now calculate the electrostatic force between the charges:
\(F = k\frac{|q_1q_2|}{r^2} = 8.99 × 10^9 \frac{|1 \times 10^{-9} \times (-2) \times 10^{-9}|}{(\sqrt{3})^2} = 3.99 \times 10^{-9} \mathrm{N}\)
05
Find the unit vector
To find the vector force acting on each charge, we need to determine the unit vector from \((0,0,0)\) to \((1,1,1)\). The unit vector can be calculated as:
\(\hat{r} = \frac{(1-0)\hat{i} + (1-0)\hat{j} + (1-0)\hat{k}}{\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)
06
Determine the vector force
Finally, we can find the vector force acting on each charge by multiplying the electrostatic force magnitude \(F\) with the unit vector:
\(F_{12} = F\hat{r} = 3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)
Since the force acting on the first charge is equal and opposite to the force acting on the second charge, we have:
\(F_{21} = -F\hat{r} = -3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)
So the vector force acting on the \(1 \mathrm{nC}\) charge at \((0,0,0)\) is \(3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\mathrm{N}\). The vector force acting on the \(-2 \mathrm{nC}\) charge at \((1,1,1)\) is \(-3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\mathrm{N}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electrostatic Force
Electrostatic force is a fundamental concept in physics, particularly in the study of charge interactions. This force arises between two charged objects. Its magnitude can be calculated using Coulomb's Law, which considers the size of each charge and the distance between them. According to Coulomb's Law, the force (\(F\)) between charges is calculated by:\[ F = k\frac{|q_1q_2|}{r^2} \]where:
- \(k\) is Coulomb's constant, approximately equal to \(8.99 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2\).
- \(q_1\) and \(q_2\) are the charges.
- \(r\) is the separation between the charges.
Vector Force Calculation
Calculating the vector nature of forces involves both magnitude and direction. This means that just knowing how strong a force is (its magnitude) isn't enough; we also need to know the direction in which it acts. To find the vector force between two charges, we apply Coulomb's Law to find the magnitude of the force first. Then, we determine the direction as indicated by a unit vector pointing from one charge to the other. Once you have the magnitude of the force (\(F\)) and the unit vector (\(\hat{r}\)), the vector force is:\[ \vec{F} = F\hat{r} \]For example, if the calculated magnitude is \(3.99 \times 10^{-9} \mathrm{N}\), and the charges are positioned from \((0,0,0)\) to \((1,1,1)\), the vector force reflects both the calculated steps and the spatial direction.
Charge Interaction
Charge interaction is the principle that describes how two charges interact with each other. There are two types of charge: positive and negative.
- Like charges repel each other.
- Opposite charges attract each other.
Unit Vector
A unit vector is a vector that has a magnitude of one and is used to indicate direction. In the context of electrostatic force, the unit vector helps to translate the scalar magnitude of a force into a vector form indicating the direction of the force. Consider charges located at points \( (0,0,0) \) and \( (1,1,1) \). The corresponding unit vector \(\hat{r}\) is calculated as:\[ \hat{r} = \frac{(1-0)\hat{i} + (1-0)\hat{j} + (1-0)\hat{k}}{\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) \]This unit vector specifies not only the direction of the force between the charges but also ensures that the magnitude of the direction does not alter the size of the calculated force. It standardizes the direction, allowing the calculated electrostatic force to remain as accurate vectors.