Problem 15
Question
Two charges, \(q_{1}\) and \(q_{2},\) are located at the origin and at \((0.50 \mathrm{~m}, 0),\) respectively. Where on the \(x\) -axis must a third charge, \(q_{3},\) of arbitrary sign be placed to be in electrostatic equilibrium if (a) \(q_{1}\) and \(q_{2}\) are like charges of equal magnitude, (b) \(q_{1}\) and \(q_{2}\) are unlike charges of equal magnitude, and (c) \(q_{1}=+3.0 \mu \mathrm{C}\) and \(q_{2}=-7.0 \mu \mathrm{C} ?\)
Step-by-Step Solution
Verified Answer
(a) 0.25 m. (b) 0.25 m. (c) No stable position on the axis.
1Step 1: Analyzing Scenario (a)
For scenario (a), since both charges \(q_1\) and \(q_2\) are like charges and have equal magnitude, the third charge \(q_3\) must be placed between them. The electrostatic force from each charge must be equal and opposite to achieve equilibrium for \(q_3\). If placed at position \(x\) on the x-axis, closer to the origin, the force from \(q_1\) being at a closer distance will be stronger than \(q_2\). Therefore, the position must be at the midpoint between the two charges for equilibrium. Thus, \(x = \frac{0.50}{2} = 0.25\, \text{m}\).
2Step 2: Analyzing Scenario (b)
In scenario (b), \(q_1\) and \(q_2\) are unlike charges of equal magnitude. A charge placed at the midpoint experiences equal and opposite forces from \(q_1\) and \(q_2\) because they attract each other. Here, the equilibrium for \(q_3\) is unstable unless precisely at the midpoint, otherwise it is drawn towards either charge. However, symmetry ensures the position of stability remains at \(x = 0.25\, \text{m}\).
3Step 3: Analyzing Scenario (c)
In scenario (c), we have \(q_1 = +3.0 \mu C\) and \(q_2 = -7.0 \mu C\). Since the charges are not equal, unlike charges attract, and now the charge \(q_3\) has different magnitudes of attraction/repulsion. The position \(x\) must satisfy the condition of equality of the forces:\[ \frac{k \cdot 3.0 \times 10^{-6}}{x^2} = \frac{k \cdot 7.0 \times 10^{-6}}{(0.50-x)^2} \]Solving for \(x\) via manipulation gives:\(\frac{3}{x^2} = \frac{7}{(0.5 - x)^2}\)Cross-multiplying and simplifying\(3(0.5-x)^2 = 7x^2\)\(3(0.25 - x + x^2) = 7x^2\)\(0.75 - 3x + 3x^2 = 7x^2\)\(4x^2 + 3x - 0.75 = 0\)Using the quadratic formula, \(x = \frac{-3 \pm \sqrt{(3^2 - 4 \times 4 \times (-0.75))}}{2 \times 4}\)\(x = \frac{-3 \pm \sqrt{33}}{8}\)Solving gives possible positions of stability \(x \approx 0.842 \text{ m}\) (which is outside the region between \(q_1\) and \(q_2\), and therefore, not valid for \(q_3\) on the x-axis). Hence, there is no stable position for \(q_3\) on the axis.
Key Concepts
Understanding Point ChargesCoulomb's Law: The Foundation of ElectrostaticsExploring Electric ForceDeciphering Charge Interaction
Understanding Point Charges
Point charges are fundamental concepts in electrostatics. They refer to electric charges that are considered to occupy a single point in space. This makes them excellent theoretical models for studying electric fields and forces because they simplify calculations.
Point charges are often used in physics problems to represent charged particles like electrons or protons. When calculating the forces and fields generated by these, we treat them as if all their charge is concentrated at a single point.
Point charges are often used in physics problems to represent charged particles like electrons or protons. When calculating the forces and fields generated by these, we treat them as if all their charge is concentrated at a single point.
- They are idealized, meaning they don't extend in space.
- They simplify calculations of electric forces and fields.
- The interaction between point charges helps us understand many phenomena in electrostatics.
Coulomb's Law: The Foundation of Electrostatics
Coulomb's Law is crucial in understanding electrostatic forces between point charges. Named after Charles-Augustin de Coulomb, this fundamental law states that the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
The law can be mathematically expressed as:\[F = k \frac{|q_1 \cdot q_2|}{r^2}\]where:
The law can be mathematically expressed as:\[F = k \frac{|q_1 \cdot q_2|}{r^2}\]where:
- \(F\) is the magnitude of the electrostatic force.
- \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\).
- \(q_1\) and \(q_2\) are the magnitudes of the two charges.
- \(r\) is the distance between the charges.
Exploring Electric Force
Electric force is the force of interaction between charged particles. According to Coulomb's Law, this force depends on the magnitude of the charges and the distance between them. It's either attractive or repulsive, defining how charged particles influence one another.
To better understand electric forces:
To better understand electric forces:
- Electric force is a vector quantity, meaning it has both a magnitude and a direction.
- When charges are like (both positive or negative), the force is repulsive.
- When charges are unlike (one positive and one negative), the force is attractive.
- The electric force acts along the line joining the centers of the two charges.
Deciphering Charge Interaction
Charge interaction is the interactive force experienced between charges. It stems from Coulomb's Law and governs how charges influence each other's motion.
There are some key points to consider:
There are some key points to consider:
- Like charges repel, and unlike charges attract.
- The strength of interaction is determined by the magnitudes of the charges and their separation distance.
- In equilibrium, all forces due to charge interaction must cancel out, resulting in no net force.
- Understanding how charges interact can help predict the behavior of electrical systems and design practical applications.
Other exercises in this chapter
Problem 13
Two charges are brought together until they are \(100 \mathrm{~cm}\) apart, causing the electric force between them to increase by a factor of exactly \(5 .\) W
View solution Problem 14
The distance between neighboring singly charged sodium and chlorine ions in crystals of table salt \((\mathrm{NaCl})\) is \(2.82 \times 10^{-10} \mathrm{~m}\).
View solution Problem 16
Two negative point charges are separated by \(10.0 \mathrm{~cm}\) and feel a mutual repulsive force of \(3.15 \mu \mathrm{N}\). The charge of one is three times
View solution Problem 17
An electron is placed on a line connecting two fixed point charges of equal charge but opposite sign. The distance between the charges is \(30.0 \mathrm{~cm}\)
View solution