/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Two point charges are located on... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two point charges are located on the \(y\)-axis as follows: charge \(q_1 = -1.50 \)nC at \(y = -\)0.600 m, and charge \(q_2 = +\)3.20 nC at the origin \((y = 0)\). What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_3 = +\)5.00 nC located at \(y = -\)0.400 m ?

Short Answer

Expert verified
The total force on charge \(q_3\) is approximately \(7.86 \times 10^{-7} \ N\) upwards.

Step by step solution

01

Understanding Charge Placement and Directions

Visualize the problem by placing the charges on the y-axis. Charge \(q_1 = -1.50\) nC is at \(y = -0.600\) m, charge \(q_2 = +3.20\) nC is at the origin \((y = 0)\), and charge \(q_3 = +5.00\) nC is at \(y = -0.400\) m. Note the distances between these charges: \(q_3\) is 0.200 m from \(q_1\), and 0.400 m from \(q_2\).
02

Applying Coulomb's Law

We use Coulomb's Law to calculate the force between two point charges: \( F = k \frac{|q_1 q_2|}{r^2} \), where \(k\) is the electrostatic constant \(8.99 \times 10^9 \ N \cdot m^2/C^2 \), and \(r\) is the distance between the charges. Calculate the force \(F_{31}\) between \(q_3\) and \(q_1\), and \(F_{32}\) between \(q_3\) and \(q_2\).
03

Calculating Force \(F_{31}\)

Using Coulomb's Law for \(q_3\) and \(q_1\): \[F_{31} = k \frac{|(-1.50\, \text{nC}) \times (+5.00\, \text{nC})|}{(0.200\, \text{m})^2} = 8.99 \times 10^9 \cdot \frac{7.50 \times 10^{-18}}{0.0400} \approx 1.686 \times 10^{-6} \ N.\] The direction is up along the y-axis because \(q_3\) is repelled by \(q_1\).
04

Calculating Force \(F_{32}\)

For \(q_3\) and \(q_2\): \[F_{32} = k \frac{|(+3.20\, \text{nC}) \times (+5.00\, \text{nC})|}{(0.400\, \text{m})^2} = 8.99 \times 10^9 \cdot \frac{16.00 \times 10^{-18}}{0.1600} \approx 8.99 \times 10^{-7}\ N.\] The direction is down the y-axis because \(q_2\) is attracting \(q_3\).
05

Finding Net Force on Charge \(q_3\)

The net force is the vector sum of \(F_{31}\) and \(F_{32}\). Since both forces are along the y-axis but in opposite directions, calculate: \[ F_{ ext{net}} = F_{31} - F_{32} \approx 1.686 \times 10^{-6} \ N - 8.99 \times 10^{-7} \ N = 7.86 \times 10^{-7} \ N. \] The net force direction is upwards because the upward force \(F_{31}\) is larger.

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 91Ó°ÊÓ!

Key Concepts

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

Electrostatic Force
Electrostatic force is the force that occurs between two electrically charged objects. It is described by Coulomb's Law, which states that the magnitude of the force between two 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 this force is: \[ F = k \frac{|q_1 q_2|}{r^2} \] where:
  • \(F\) is the electrostatic force,
  • \(k\) is the electrostatic constant \(8.99 \times 10^9 \ N \cdot m^2/C^2\),
  • \(q_1\) and \(q_2\) are the point charges,
  • \(r\) is the distance between the charges.

This force can be attractive or repulsive depending on the nature of the charges involved. If the charges are of opposite signs, the force is attractive; if they are of the same sign, the force is repulsive. Understanding and calculating this force allows us to comprehend how charged particles interact in a static electric field.
Point Charges
Point charges refer to objects that have a very small size compared to their separation distance, allowing their charge to be treated as if it were concentrated at a single point. This simplification is essential when applying Coulomb's Law.
Point charges are fundamental in electrostatics because they create electric fields and interact via electrostatic forces. Two point charges can influence each other even when they are not in direct contact.
In the exercise, each of the charges—\(q_1 = -1.50\) nC, \(q_2 = +3.20\) nC, and \(q_3 = +5.00\) nC—is treated as a point charge. Their positions on the y-axis simplify calculations to one dimension, making it easier to apply the principle of superposition to find vector sums and net forces.
Vector Sum
The concept of a vector sum is essential when dealing with forces in physics. Forces are vector quantities, meaning they have both magnitude and direction.
When multiple forces act on an object, the net force is determined by finding the vector sum of these individual forces. In the case of electrostatic forces between point charges, the vector sum accounts for the directionality of each force involved. This is important because different forces may not act in the same direction.
In the provided exercise, calculating the net force on charge \(q_3\) involves the vector sum of forces \(F_{31}\) and \(F_{32}\). Since these forces are along the same line but in opposite directions, their net effect is calculated by taking the difference, acknowledging that force direction is crucial to understanding resultant impacts on charge trajectories.
Electric Charge Interaction
Electric charge interaction is the basis for explaining how charged objects exert forces upon each other. This interaction is insightful when analyzing the behavior of like and unlike charges in close proximity.
In the exercise, the interactions manifest as forces exerted by two point charges on a third charge. Here's how they interact:
  • Charge \(q_1 = -1.50\) nC repels \(q_3 = +5.00\) nC because opposite charges attract, creating a force \(F_{31}\) upwards on the y-axis.
  • Charge \(q_2 = +3.20\) nC attracts \(q_3\), resulting in a force \(F_{32}\) downwards, due to the like charge interaction being repulsive for other potential configurations.

This exercise helps visualize charge interaction principles, providing insights into the intricacies of electrostatic force calculations and their real-world implications. Understanding this concept is crucial for anyone studying electric fields and charges.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Four identical charges \(Q\) are placed at the corners of a square of side \(L\). (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

Two thin rods of length \(L\) lie along the \(x\)-axis, one between \(x = \frac{1}{2} a\) and \(x = \frac{1}{2} a + L\) and the other between \(x = -\frac{1}{2} a\) and \(x = -\frac{1}{2} a - L\). Each rod has positive charge \(Q\) distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive x-axis. (b) Show that the magnitude of the force that one rod exerts on the other is $$F = {Q^2 \over 4\pi\epsilon_0 L^2} ln [ {(a + L)^2 \over a(a + 2L)} ]$$ (c) Show that if \(a\) \(\gg\) \(L\), the magnitude of this force reduces to \(F = Q^2/4\pi\epsilon_0 a^2\). (\(Hint\): Use the expansion ln \((1 + z) = z - \frac{1}{2} z^2 + \frac{1}{3} z^3 - \cdot\cdot\cdot\), valid for \(\mid z \mid\ll1\). Carry \(all\) expansions to at least order \(L^2/a^2.\)) Interpret this result.

In a follow-up experiment, a charge of \(+40\) pC was placed at the center of an artificial flower at the end of a 30-cm long stem. Bees were observed to approach no closer than 15 cm from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \space N/C; (b) 16 \space N/C; (c) 2.7 \times 10^{-10} \space N/C; (d) 4.8 \times 10^{-10} \space N/C.\)

A charge \(q_1 = +\)5.00 nC is placed at the origin of an \(xy\)-coordinate system, and a charge \(q_2 = -\)2.00 nC is placed on the positive \(x\)-axis at \(x = \)4.00 cm. (a) If a third charge \(q_3 = +\)6.00 nC is now placed at the point \(x =\) 4.00 cm, \(y =\) 3.00 cm, find the \(x\)- and \(y\)-components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.