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Chapter 17: Problem 22

\(\bullet$$\bullet\) Two point charges are placed on the \(x\) axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m} .\) What are the magnitude and direction of the net force exerted by these two charges on a negative point charge \(q_{3}=-0.600 \mathrm{nC}\) placed at the origin?

Short Answer

Expert verified
The magnitude of the net force is \(2.393 \times 10^{-7} \,\text{N}\), directed towards \(q_1\) (positive \(x\) axis).

Step by step solution

01

Identify the Charges and Positions

We have three point charges. Charge \( q_1 = +4.00 \,\text{nC} \) is placed at \( x = 0.200 \,\text{m} \), charge \( q_2 = +5.00 \,\text{nC} \) is at \( x = -0.300 \,\text{m} \), and the charge \( q_3 = -0.600 \,\text{nC} \) is at the origin \( x = 0.000 \,\text{m} \).
02

Calculate the Force between q1 and q3

Use Coulomb's Law to find the force between charges \( q_1 \) and \( q_3 \). The formula is \( F = \frac{{k |q_1 q_3|}}{r_{13}^2} \), where \( k = 8.99 \times 10^9 \,\text{Nm}^2/\text{C}^2 \) and \( r_{13} = 0.200 \,\text{m} \).\[F_{13} = \frac{(8.99 \times 10^9) (4.00 \times 10^{-9}) (0.600 \times 10^{-9})}{(0.200)^2} = 5.39 \times 10^{-7} \,\text{N}\]The force \( F_{13} \) is attractive because \( q_3 \) is negative and is directed towards \( q_1 \).
03

Calculate the Force between q2 and q3

Similarly, calculate the force exerted by \( q_2 \) on \( q_3 \) using the same formula. Here \( r_{23} = 0.300 \,\text{m} \).\[F_{23} = \frac{(8.99 \times 10^9) (5.00 \times 10^{-9}) (0.600 \times 10^{-9})}{(0.300)^2} = 2.997 \times 10^{-7} \,\text{N}\]The force \( F_{23} \) is also attractive because \( q_3 \) is negative and is directed towards \( q_2 \).
04

Determine the Direction of Each Force

\( F_{13} \) is directed towards the positive \( x \) axis, while \( F_{23} \) is directed towards the negative \( x \) axis, as both forces are attractive and \( q_3 \) is at the origin.
05

Calculate the Net Force on q3

Since \( F_{13} \) and \( F_{23} \) are in opposite directions, we subtract them to find the net force. \[ F_{\text{net}} = F_{13} - F_{23} = 5.39 \times 10^{-7} \,\text{N} - 2.997 \times 10^{-7} \,\text{N} = 2.393 \times 10^{-7} \,\text{N} \]This resultant force is directed towards \( q_1 \) as \( F_{13} > F_{23} \).

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Key Concepts

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

Electric Force
Electric force is a fundamental concept in physics, describing the attraction or repulsion between charged particles. It is governed by Coulomb's Law. This law states that the electric force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

The formula for electric force using Coulomb's Law is given by:
  • \( F = \frac{k |q_1 q_2|}{r^2} \)
Where:
  • F is the electric force,
  • k is Coulomb's constant \(8.99 \times 10^9 \,\text{Nm}^2/\text{C}^2\),
  • 辩鈧� and 辩鈧� are the magnitudes of the charges,
  • r is the distance between the charges.
The direction of the force is attractive if the charges are of opposite signs and repulsive if they are of the same sign. This principle plays a vital role in understanding how forces act on charged particles, such as in our original exercise scenario.
Point Charges
Point charges are idealized charged particles that are considered to have no size, only possessing charge. In most physics problems, real objects are approximated as point charges to simplify calculations.

In our example, two point charges, \( q_1 = +4.00 \,\text{nC} \) and \( q_2 = +5.00 \,\text{nC} \), exert forces on a third point charge \( q_3 = -0.600 \,\text{nC} \) located at the origin. Despite their idealization, point charges effectively demonstrate electrical interactions that occur in real life. By calculating the forces these charges exert on each other, we gain insights into how charge distribution affects electric fields and resulting forces.
Vector Addition
Vector addition is a critical method used to determine the resultant of two or more vectors. In the context of electric forces, it's essential because forces are vector quantities, having both magnitude and direction.

For the problem, it鈥檚 necessary to calculate the individual forces exerted by \( q_1 \) and \( q_2 \) on \( q_3 \). These forces act along the x-axis:
  • \( F_{13} \) is directed from \( q_3 \) towards \( q_1 \) on the positive x-axis.
  • \( F_{23} \) is directed from \( q_3 \) towards \( q_2 \) on the negative x-axis.
By using vector addition, these opposing forces are combined by subtracting their magnitudes to find the net force on \( q_3 \). This process highlights the importance of considering both direction and magnitude in solving physics problems involving vectors.
Electric Field
The electric field is a vector field around a charged particle that represents the force exerted per unit charge at any given point in space. It provides a way to describe the influence a charge has on its surrounding environment.

The electric field \( E \) is defined by:
  • \( E = \frac{F}{q} \)
Where:
  • F is the force experienced by a test charge,
  • q is the magnitude of the test charge.
In the context of multiple point charges, the total electric field at any point is the vector sum of the electric fields due to each charge. Related to our exercise, considering the electric fields created by \( q_1 \) and \( q_2 \) could further enhance understanding, although it's not needed to determine just the net force on \( q_3 \). Understanding electric fields allows us to predict how an additional charge would behave when introduced into a charged environment.

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Most popular questions from this chapter

\(\bullet\) . Electric field of axons. A nerve signal is transmitted through a neuron when an excess of \(\mathrm{Na}^{+}\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately 10.0\(\mu \mathrm{m}\) in diameter, and meas- urements show that about \(5.6 \times 10^{11} \mathrm{Na}^{+}\) ions per meter (each of charge \(+e )\) enter during this process. Although the axon is a long cylinder, the charge does not all enter every- where at the same time. A plausible model would be a a series of nearly point charges moving along the axon. Let us look at a 0.10 mm length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.10 \(\mathrm{mm}\) length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 \(\mathrm{cm}\) below the skin? (c) Certain sharks can respond to electric fields as weak as 1.0\(\mu \mathrm{N} / \mathrm{C}\) . How far from this segment of axon could a shark be and still detect its electric field?

\(\bullet$$\bullet\) Three point charges are arranged along the \(x\) axis. Charge \(q_{1}=-4.50 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+2.50 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m} .\) A positive point charge \(q_{3}\) is located at the origin. (a) What must the value of \(q_{3}\) be for the net force on this point charge to have magnitude 4.00\(\mu \mathrm{N} ?\) (b) What is the direction of the net force on \(q_{3} ?\) (c) Where along the \(x\) axis can \(q_{3}\) be placed and the net force on it be zero, other than the trivial answers of \(x=+\infty\) and \(x=-\infty\) ?

\(\bullet$$\bullet\) An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 \(\mathrm{m}\) in the first 3.00\(\mu\) s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignor- ing the effects of gravity? Justify your answer quantitatively.

\(\bullet\) During a violent electrical storm, a car is struck by a falling high-voltage wire that puts an excess charge of \(-850 \mu C\) on the metal car. (a) How much of this charge is on the inner sur- face of the car? (b) How much is on the outer surface?

\(\bullet$$\bullet\) In a certain region of space, the electric field \(E\) is uniform; i.e., neither its direction nor its magnitude changes in the region. (a) Use Gauss's law to prove that this region of space must be electrically neutral; that is, there must be no charge in this region. (b) Is the converse true? That is, in a region of space where there is no charge, must \(\vec{E}\) be uniform? Explain.
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