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A negative charge of \(-0.550 \space \mu\)C exerts an upward 0.600-N force on an unknown charge that is located 0.300 m directly below the first charge. What are (a) the value of the unknown charge (magnitude and sign); (b) the magnitude and direction of the force that the unknown charge exerts on the \(-\)0.550-\(\mu\)C charge?

Short Answer

Expert verified
(a) \(-0.100 \mu C\); (b) 0.600 N, downward.

Step by step solution

01

Understanding the Problem

We have two point charges: one is known with a charge of \(-0.550 \mu C\) exerting a force of 0.600 N on an unknown charge located 0.300 m below. The force is due to the electric field interaction of the two charges. We need to find the magnitude and direction of the unknown charge, as well as the force exerted by the unknown charge on the known charge.
02

Applying Coulomb's Law for Part (a)

Coulomb's law states that the magnitude of the force between two point charges can be calculated by \(F = \frac{k |q_1 q_2|}{r^2}\), where \(F\) is the force, \(k\) is Coulomb's constant \(8.99 \times 10^9 \; N \, m^2/C^2\), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the separation between the charges. Here, \(F = 0.600\, N\), \(q_1 = -0.550 \mu C\), and \(r = 0.300\, m\). We solve for \(q_2\) the unknown charge.
03

Solving for the Unknown Charge

Rearrange Coulomb's Law for \(q_2\): \[q_2 = \frac{F \cdot r^2}{k \cdot |q_1|}\]. Substitute the values: \[q_2 = \frac{0.600 \cdot (0.300)^2}{8.99 \times 10^9 \cdot 0.550 \times 10^{-6}}\]. Calculations yield \(q_2 = -0.100 \mu C\). The negative sign indicates the charge is also negative.
04

Interpreting Part (a) Results

The magnitude of the unknown charge is \(0.100 \mu C\) and it is negative. This means the force is attractive, attracting the charges toward one another, as they are of like signs.
05

Calculating Force by Unknown Charge for Part (b)

Using Newton's third law, the magnitude of the force exerted by the unknown charge on the known charge is the same as the force the known charge exerts on the unknown charge, which is 0.600 N. The direction of this force is downward, towards the unknown charge.
06

Summarizing Part (b) Results

The force exerted by the unknown charge on the \(-0.550 \mu C\) charge is 0.600 N directed downward. This is consistent with Newton's third law of motion where forces are equal in magnitude and opposite in direction.

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Key Concepts

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

Electric Force
The electric force is a fundamental interaction that arises from charged particles. It's one of the essential concepts in physics, particularly when it comes to understanding the behavior of charged particles and their interactions.
Charged particles exert forces on each other, and these forces can either attract or repel the charges. In our exercise, we observe a force of 0.600 N acting due to the interaction between two charges. Applying Coulomb's Law helps quantify this force. Coulomb's Law explicitly states the force between two point charges as:
\[F = \frac{k |q_1 q_2|}{r^2}\]
where:
  • \(F\) is the electric force between the charges.

  • \(k\) is Coulomb's constant, \(8.99 \times 10^9 \, N \, m^2/C^2\).

  • \(q_1\) and \(q_2\) are the magnitudes of the charges.

  • \(r\) is the distance between the charges.
The 0.600 N force exerted by the known charge on the unknown charge demonstrates a classic example of electric force in action, dictated by their respective magnitudes and the distance between them.
Point Charges
Point charges are hypothetical charged particles that occupy no space but have charge and mass. They're used to simplify complex problems by focusing only on the charges without considering their size or the surface level interactions.
In this exercise, we treat the charges as point charges. This allows us to apply Coulomb's Law straightforwardly, without accounting for additional factors like the distribution of charge across a surface.
By considering charges as point charges, we focus on the essential attributes needed to compute the force. In our specific task, the known charge is \(-0.550 \mu C\), and the unknown charge, discovered to be \(-0.100 \mu C\), both treated as point charges. This simplification helps highlight the core principles of electric interactions, making calculations more manageable for foundational learning.
Newton's Third Law
Newton's Third Law of Motion is a pivotal principle in understanding the interactions between objects, including point charges in electromagnetism. This law states that for every action, there is an equal and opposite reaction.
In the context of our exercise, this means that if the -0.550 \(\mu C\) charge exerts a force of 0.600 N upward on the unknown charge, then the unknown charge must exert an equal force of 0.600 N downward on the known charge.
This concept confirms the mutual interactions between the charges, ensuring consistency in the forces exerted. Despite the complexity of electric charge interactions, Newton's Third Law provides clarity by assuring that the forces are balanced. So in part (b) of the exercise, we straightforwardly concluded that the unknown charge also exerts 0.600 N, directed downwards, is consistent with Newton’s law.

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Most popular questions from this chapter

The earth has a downward-directed electric field near its surface of about 150 N\(/\)C. If a raindrop with a diameter of 0.020 mm is suspended, motionless, in this field, how many excess electrons must it have on its surface?

An average human weighs about 650 N. If each of two average humans could carry 1.0 C of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650-N weight?

Point charges \(q_1 = -\)4.5 nC and \(q_2 = +\)4.5 nC are separated by 3.1 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of 36.9\(^\circ\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9}\) N \(\cdot\) m ?

A straight, nonconducting plastic wire 8.50 cm long carries a charge density of \(+\)175 nC\(/\)m distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 cm directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 cm directly above its center.

A \(+\)2.00-nC point charge is at the origin, and a second \(-\)5.00-nC point charge is on the \(x\)-axis at \(x = \)0.800 m. (a) Find the electric field (magnitude and direction) at each of the following points on the x-axis: (i) \(x =\) 0.200 m; (ii) \(x =\) 1.20 m; (iii) \(x = -\)0.200 m. (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

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