Chapter 21: Problem 67
Two particles having charges \(q_1 =\) 0.500 nC and \(q_2 =\) 8.00 nC are separated by a distance of 1.20 m. At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?
Short Answer
Expert verified
The electric field is zero at approximately 0.421 m from the 0.500 nC charge.
Step by step solution
01
Understand the Problem
We need to find a point along the line connecting two charges, where the electric fields due to each charge cancel out, resulting in a net electric field of zero.
02
Use the Formula for Electric Field
We use the formula for the electric field due to a point charge, which is given by \[ E = \frac{k \cdot |q|}{r^2} \]where \( E \) is the electric field, \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point of interest.
03
Define the Point of Interest
Let the point along the line connecting the two charges be at a distance \( x \) from the charge \( q_1 \). Thus, the distance from this point to \( q_2 \) will be \( 1.20 - x \).
04
Set the Electric Fields Equal and Opposite
For the electric field to be zero at some point, the magnitude of the electric field due to \( q_1 \) must equal the magnitude of the electric field due to \( q_2 \) but in opposite directions:\[ \frac{k \cdot q_1}{x^2} = \frac{k \cdot q_2}{(1.20 - x)^2} \]
05
Simplify the Equation
Cancel \( k \) from both sides:\[ \frac{q_1}{x^2} = \frac{q_2}{(1.20 - x)^2} \]
06
Solve for x
Substituting \( q_1 = 0.500 \times 10^{-9} C \) and \( q_2 = 8.00 \times 10^{-9} C \), we have:\[ \frac{0.500}{x^2} = \frac{8.00}{(1.20 - x)^2} \]Cross-multiply to get:\[ 0.500 \cdot (1.20 - x)^2 = 8.00 \cdot x^2 \]Solve this quadratic equation for \( x \).
07
Solving the Quadratic Equation
Expanding the equation:\[ 0.500 \cdot (1.44 - 2.4x + x^2) = 8.00 \cdot x^2 \]\[ 0.720 - 1.20x + 0.500x^2 = 8.00x^2 \]Collect all terms on one side:\[ 0.720 - 1.20x - 7.50x^2 = 0 \]Rearrange to form a standard quadratic equation:\[ 7.50x^2 + 1.20x - 0.720 = 0 \]Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 7.50,\ b = 1.20,\ c = -0.720 \).
08
Calculate and Interpret Result
Solving using the quadratic formula:\[ x = \frac{-1.20 \pm \sqrt{(1.20)^2 - 4 \cdot 7.50 \cdot (-0.720)}}{2 \cdot 7.50} \]Calculate the discriminant and find the roots. The physically meaningful root is where the distance from the charge is positive and less than 1.20 meters.
09
Complete the Solution
Solve for the roots of the quadratic equation. Use only the root that is physically meaningful (closer to \( q_1 \) if it results in a positive x under 1.20 meters).One possible outcome could yield \( x \approx 0.421 \text{ m} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
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 in physics that describes the force between two charged objects. The law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between their centers. This can be expressed mathematically as:
- \[ F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \]
- \( F \) is the force between the charges.
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
- \( r \) is the distance between the centers of the two charges.
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2 \).
Quadratic Equation in Physics
In the realm of physics, quadratic equations often arise in problems involving motion, forces, and fields. Solving a quadratic equation is essential when variables in a squared relationship describe physical phenomena. The standard form of a quadratic equation is:
- \[ ax^2 + bx + c = 0 \]
- \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Electric Field of Point Charges
The electric field at a point in space due to a point charge is another critical concept in understanding electric forces. The field describes how an electric force would exert on a test charge placed at a specific location. The electric field \( E \) due to a point charge can be calculated using the formula:
- \[ E = \frac{k \, |q|}{r^2} \]
- \( E \) is the electric field strength.
- \( q \) is the charge creating the field.
- \( r \) is the distance from the charge where the field is measured.
- \( k \) is Coulomb's constant.