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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 that of the other. (a) How much charge does each have? (b) What would be the force if the total charge were instead equally distributed on both point charges?

Short Answer

Expert verified
\(q_1 = -3.45 \mu C, q_2 = -1.15 \mu C\); equal distribution force is 1.83 µN.

Step by step solution

01

Understanding Coulomb's Law

To find the charges, we need to use Coulomb's Law which states that the force between two point charges is given by \[ F = k \frac{|q_1 q_2|}{r^2} \]where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ N m²/C²}) \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the separation distance between them.
02

Setting Up Relationships and Equations

We are given that the charge of one point (say, \( q_1 \)) is three times that of the other \( q_2 \), so we can write:\[ q_1 = 3q_2 \]We also know the distance \( r = 10.0 \text{ cm} = 0.1 \text{ m} \) and the force \( F = 3.15 \mu \text{N} = 3.15 \times 10^{-6} \text{ N} \). Substitute these into Coulomb's Law:\[ 3.15 \times 10^{-6} = 8.99 \times 10^9 \frac{3q_2^2}{0.1^2} \]
03

Solving for the Charges

Rearrange the equation from Step 2 to solve for \( q_2^2 \): \[ 3.15 \times 10^{-6} = 8.99 \times 10^9 \cdot 300 q_2^2 \]\[ q_2^2 = \frac{3.15 \times 10^{-6}}{2.697 \times 10^9} \]Calculate \( q_2 \) and \( q_1 = 3q_2 \).
04

Solving for Equal Charge Distribution

For part (b), if the total charge \( Q = q_1 + q_2 \) is equally distributed such that each charge \( q = \frac{Q}{2} \), plug \( q = \frac{q_1 + q_2}{2} \) back into Coulomb's Law:\[ F = k \cdot \frac{q^2}{r^2} \].Calculate the new force.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Force
Electric force refers to the attractive or repulsive interaction between any two charged objects. This concept is fundamental to understanding Coulomb's Law, which describes how the force depends on the magnitudes of the charges and the distance between them.
The electric force is mathematically represented by the formula \( F = k \frac{|q_1 q_2|}{r^2} \), where:
  • \( F \) is the force between the charges.
  • \( k \) is Coulomb's constant \((8.99 \times 10^9 \text{ N m²/C²})\).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance separating the charges.
In this exercise, we analyze how electric forces are responsible for the interactions between two point charges. They push each other with a repulsive force, as both charges are negative.
This repulsive nature arises because like charges repel, opposing each other with a calculated force through the principles of electric force.
Point Charges
Point charges refer to the idealized representation of charged objects in physics. This concept assumes charges to be concentrated at a single point in space, simplifying the calculation of electric forces.
Point charges are crucial in applying Coulomb's Law effectively since it helps to focus on calculating the impact of charges without considering the physical size and shape. In this exercise, the charges are negatively charged point charges separated by 10 cm. By treating these as point charges, we can easily compute the forces acting on them using Coulomb’s formula. This allows us to concentrate solely on the interactions caused by the charges themselves, shedding complexities tied to real-world dimensions and charge distributions.
Charge Distribution
Charge distribution plays a critical role in determining electric forces. It denotes how electric charge is spread across a given object or system. In problems involving point charges, charge distribution can influence the net forces experienced by the charges.
In our exercise, each charge was considered as a point with a specific numerical value, and the relationship was established such that one charge is three times the other. This uneven charge distribution influences both the initial and subsequent interactions between the charges. When charges are uniformly distributed, as in the alternate scenario where the total charge is equally split between two charges, it alters the force calculated by Coulomb's Law. This demonstrated that proper knowledge of charge distribution affects the outcome regarding how forces interact and manifest between various charges in a given environment.
Repulsive Force
Repulsive force arises between two like charges, meaning elements that have similar kinds of electric charge, such as two positive or two negative charges. In this particular exercise, the repulsive force is of primary interest because both charges involved are negative. Since like charges experience a repulsive interaction, it causes the charges to move apart from each other.
The strength of this repulsive force depends on the amount of charge each point charge holds and their separation distance. Using Coulomb's Law, this interaction is quantified to show how two negative charges repel each other with a force specified in micro-Newtons. Repulsive force has real-world implications, influencing how charge-based systems behave, from atomic particles to larger scale electrical applications, underscoring the practical importance of understanding these fundamental principles.

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Most popular questions from this chapter

A particle with a mass of \(2.0 \times 10^{-5} \mathrm{~kg}\) and a charge of \(+2.0 \mu \mathrm{C}\) is released in a (parallel plate) uniform horizontal electric field of \(12 \mathrm{~N} / \mathrm{C}\). (a) How far horizontally does the particle travel in \(0.50 \mathrm{~s} ?\) (b) What is the horizontal component of its velocity at that point? (c) If the plates are \(5.0 \mathrm{~cm}\) on each side, how much charge is on each?

What would be the magnitude and the direction of an electric field that would just support the weight of a proton near the surface of the Earth? What about an electron?

A negatively charged pith ball (mass \(6.00 \times 10^{-3} \mathrm{~g}\), charge \(-1.50 \mathrm{nC}\) ) is suspended vertically from a light nonconducting string of length \(15.5 \mathrm{~cm} .\) This apparatus is then placed in a horizontal uniform electric field. After being released, the pith ball comes to a stable position at an angle of \(12.3^{\circ}\) to the left of the vertical. (a) What is the direction of the external electric field? (b) Determine the magnitude of the electric field.

An electrically neutral thin, square metal slab, measuring \(5.00 \mathrm{~cm}\) on a side, is placed in a uniform external field that is perpendicular to its square area. (a) If the top of the slab becomes negatively charged, what is the direction of the external field? (b) If the external field strength is \(1250 \mathrm{~N} / \mathrm{C}\), what are the direction and strength of the field that is generated by the charges induced on the slab? (c) What is the total charge on the negative side of the slab?

What are the magnitude and direction of the electric field at a point \(0.75 \mathrm{~cm}\) away from a point charge of \(+2.0 \mathrm{pC} ?\)

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