Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 17: Problem 43
\(\bullet\) \(\bullet\) Two particles having charges of \(+0.500 \mathrm{nC}\) and \(+8.00 \mathrm{nC}\) are separated by a distance of 1.20 \(\mathrm{m}\) . (a) At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero? (b) Where would the net electric field be zero if one of the charges were negative?
Short Answer
Expert verified
(a) 0.40 m from the +0.500 nC charge; (b) beyond the 1.20 m point.
Step by step solution
01
Understand the System
Visualize the system as two point charges located along a straight line, with charge \( q_1 = +0.500 \, \mathrm{nC} \) and charge \( q_2 = +8.00 \, \mathrm{nC} \), separated by \( 1.20 \, \mathrm{m} \). We need to find the position along this line where their electric fields cancel each other.
02
Set Up the Equation for Equal Electric Fields
The electric field due to a point charge is given by \( E = \frac{k \cdot q}{r^2} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \mathrm{N m^2/C^2} \). For the net field to be zero, the fields due to \( q_1 \) and \( q_2 \) must be equal in magnitude but opposite in direction:\[ \frac{k \cdot q_1}{x^2} = \frac{k \cdot q_2}{(1.20 - x)^2} \] where \( x \) is the distance from \( q_1 \).
03
Simplify and Solve the Equation
Cancel \( k \) from both sides and rearrange terms:\[ \frac{q_1}{x^2} = \frac{q_2}{(1.20 - x)^2} \]Substitute \( q_1 = 0.500 \, \mathrm{nC} \) and \( q_2 = 8.00 \, \mathrm{nC} \) to get:\[ \frac{0.500}{x^2} = \frac{8.00}{(1.20 - x)^2} \]Solving for \( x \), square both sides and simplify: \[ 0.500 (1.20 - x)^2 = 8.00 x^2 \]\[ 0.500 \cdot (1.44 - 2.40x + x^2) = 8.00x^2 \]\[ 0.720 - 1.20x + 0.500x^2 = 8.00x^2 \]\[ 0 = 7.50x^2 + 1.20x - 0.720 \]Use the quadratic formula to solve for \( x \):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 7.50, b = -1.20, c = -0.720 \).
04
Calculate the Roots of the Quadratic Equation
Plug the values into the quadratic formula:\[ x = \frac{1.20 \pm \sqrt{(-1.20)^2 - 4 \cdot 7.50 \cdot (-0.720)}}{2 \cdot 7.50} \]Evaluate the discriminant:\[ (-1.20)^2 - 4 \cdot 7.50 \cdot (-0.720) = 1.44 + 21.60 = 23.04 \]\[ x = \frac{1.20 \pm \sqrt{23.04}}{15.00} \]\[ \sqrt{23.04} = 4.80 \]\[ x = \frac{1.20 \pm 4.80}{15.00} \]Find two solutions: one will be negative and one positive, only the positive solution is meaningful:\[ x = \frac{6.00}{15.00} = 0.40 \, \mathrm{m} \]
05
Consider the Case with one Negative Charge
If one charge is negative, the net electric field zero point will be located outside the segment connecting the charges. Assume \( q_1 = +0.500 \, \mathrm{nC} \) and \( q_2 = -8.00 \, \mathrm{nC} \). Set up the equation:\[ \frac{k \cdot q_1}{x^2} = \frac{k \cdot |q_2|}{(1.20 + x)^2} \]Simplify with \( x \) as the distance from \( q_1 \) on the opposite direction:\[ \frac{0.500}{x^2} = \frac{8.00}{(1.20 + x)^2} \]Solve similarly to Step 3, but since it involves cubic equations or trial/error methods, more advanced mathematics or numerical approximation might be needed for an exact location. The solution will be beyond the 1.20 m location.
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Point Charge
In physics, a point charge is an idealized or simplified model used to represent an electric charge as being located at a single point in space. This is a useful concept when solving problems related to electric fields and forces. Despite its simplicity, the point charge concept allows us to calculate the effects of an electric charge in a precise way.
- When we refer to a point charge, we imagine the entire charge being concentrated at a single point with no physical size.
- This model is particularly effective for analyzing electric fields and forces at distances much larger than the size of the charged object.
- In the original problem, both charges are considered point charges, enabling us to apply simple equations to determine the electric properties along the line connecting them.
Coulomb's Law
Coulomb's law is a fundamental principle in physics that describes the force exerted between two point charges. It's analogous to Newton's law of universal gravitation but applies to electric charges instead of masses.
- The law states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
- The formula is given by: \[ F = k \frac{{|q_1 \cdot q_2|}}{{r^2}} \]where: - \( F \) is the force between the charges, - \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \mathrm{N m^2/C^2} \), - \( q_1 \) and \( q_2 \) are the amounts of the charges, - \( r \) is the distance between the centers of the two charges.
- This interaction can be attractive or repulsive depending on the nature of the charges (like charges repel, unlike charges attract).
Quadratic Formula
The quadratic formula is an indispensable tool in algebra, employed to find the roots of quadratic equations, which are polynomials of degree two. It provides an exact method for finding the values where the function equals zero.
- A standard quadratic equation is represented as: \[ ax^2 + bx + c = 0 \]
- The quadratic formula for solving this equation is: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
- Here, \( a \), \( b \), and \( c \) are constants where \( a eq 0 \), and \( x \) represents the variable or solution.
- Under the square root, \( b^2 - 4ac \) is known as the discriminant, which determines the nature of the roots: - If positive, there are two real and distinct solutions. - If zero, there is exactly one real solution. - If negative, the solutions are complex.
Electric Field Cancellation
Electric field cancellation occurs when two or more electric fields combine to yield a net electric field of zero at a particular point in space. This concept is vital when studying interactions between charges.
- An electric field is the region around a charged object where the force of electricity can be felt by other charges.
- Cancellation happens when electric fields have equal magnitudes but opposite directions, effectively neutralizing each other.
- In the given exercise, we aimed to find the position along the line connecting two charges where their electric fields cancelled each other completely:
- For two positive charges, this point lies between them due to the repelling forces.
- However, with one charge negative (attractive force), the cancellation point would be outside the region between the charges.
- This principle is used widely in balancing forces in electric circuits and other electric field applications.
