/*! 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 66 A charge \(q_{1}=+5.00 \mathrm{n... [FREE SOLUTION] | 91影视

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Chapter 21: Problem 66

A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\) -coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\) -axis at \(x=4.00 \mathrm{cm} .\) (a) If a third charge \(q_{3}=\) \(+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{cm}, y=3.00 \mathrm{cm}\) 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.

Short Answer

Expert verified
The force on \(q_3\) is 10.27 N at \(-32.5^\circ\) from the positive \(x\)-axis.

Step by step solution

01

Calculate the distance between charges

First, calculate the distance between charge \(q_1\) at the origin and charge \(q_3\) at the point \((4.00 \, \text{cm}, 3.00 \, \text{cm})\). Use the distance formula which gives \(r_{13} = \sqrt{(4.00)^2 + (3.00)^2} = 5.00 \, \text{cm}\).
02

Determine direction vectors

Determine the unit vector components from charge \(q_1\) to charge \(q_3\), which has components: \(\hat{i}_{13} = \frac{4}{5}\) and \(\hat{j}_{13} = \frac{3}{5}\). Similarly, for charge \(q_2\) to \(q_3\), the unit vector is purely in the \(y\)-direction, \(\hat{j}_{23} = 1\), since \(q_2\) and \(q_3\) are vertically aligned.
03

Calculate forces using Coulomb's Law

Apply Coulomb's law \(F = \frac{k \cdot |q_1q_3|}{r_{13}^2}\) for both pairs of charges \(q_1, q_3\) and \(q_2, q_3\).For \(q_1, q_3\): \(F_{13} = \frac{(8.99 \times 10^9) \cdot (5.00 \times 10^{-9})(6.00 \times 10^{-9})}{(0.05)^2} = 10.79 \, \text{N}\).For \(q_2, q_3\): \(F_{23} = \frac{(8.99 \times 10^9) \cdot (2.00 \times 10^{-9})(6.00 \times 10^{-9})}{(0.03)^2} = 11.99 \, \text{N}\).
04

Resolve forces into components

Resolve each force into its components. For \(F_{13}\), using the unit vector components: \(F_{13x} = 10.79 \times \frac{4}{5} = 8.63 \, \text{N}\) and \(F_{13y} = 10.79 \times \frac{3}{5} = 6.47 \, \text{N}\).For \(F_{23}\), since it only acts in the \(y\)-direction: \(F_{23x} = 0\) and \(F_{23y} = 11.99 \, \text{N}\). Since \(q_2\) is negative, the direction is reversed.
05

Calculate total force components

Add the components of the forces to find the total force on \(q_3\):\(F_{3x} = F_{13x} + F_{23x} = 8.63 + 0 = 8.63 \, \text{N}\).\(F_{3y} = F_{13y} - F_{23y} = 6.47 - 11.99 = -5.52 \, \text{N}\).
06

Calculate magnitude and direction of total force

Find the magnitude of the total force using Pythagorean theorem:\[ F = \sqrt{(8.63)^2 + (-5.52)^2} = 10.27 \, \text{N} \]Find the direction by calculating the angle \(\theta\) from the positive \(x\)-axis with \(\tan \theta = \frac{-5.52}{8.63}\), giving an angle of \(\theta = \arctan\left(\frac{-5.52}{8.63}\right) = -32.5^\circ\).

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. Charges are typically described as positive or negative, like in this exercise where we have three charges: \( q_1 = +5.00 \, ext{nC} \), \( q_2 = -2.00 \, ext{nC} \), and \( q_3 = +6.00 \, ext{nC} \). Charged objects interact with each other through the electrostatic force, which is governed by Coulomb's Law.

The interaction between charges depends on their magnitude and sign. Like charges repel each other, while opposite charges attract. For example, \( q_1 \) and \( q_3 \) have like charges (+), leading to repulsion, whereas \( q_2 \) and \( q_3 \) have opposite charges, causing attraction. Understanding the nature of charges is crucial for analyzing the forces at play.
Vector Components
In physics, vectors are quantities that have both magnitude and direction. For the problem at hand, we must resolve the forces between charges into their vector components to understand how they affect \( q_3 \). Each force can be broken into parts along the \( x \)- and \( y \)-axis.

Determining these components involves using unit vectors and trigonometric relationships. For instance, the force \( F_{13} \) between \( q_1 \) and \( q_3 \) needs to be split into \( F_{13x} \) and \( F_{13y} \). We use unit vector properties, where \( \hat{i}_{13} = \frac{4}{5} \) and \( \hat{j}_{13} = \frac{3}{5} \), to calculate these. This breakdown is essential for adding forces adequately. The sum of the \( x \)-components gives the total force in the \( x \)-direction, and similarly for the \( y \)-components.
Force Magnitude and Direction
After resolving the components of each force, the next task is to determine the overall effect on \( q_3 \) by calculating the force's magnitude and direction. The total force components we found are \( F_{3x} = 8.63 \, \text{N} \) and \( F_{3y} = -5.52 \, \text{N} \).

The magnitude of the force is calculated using the Pythagorean theorem: \[ F = \sqrt{(F_{3x})^2 + (F_{3y})^2} = \sqrt{(8.63)^2 + (-5.52)^2} = 10.27 \, \text{N} \] The direction can be found using trigonometry with the tangent function: \[ \tan \theta = \frac{F_{3y}}{F_{3x}} = \frac{-5.52}{8.63} \] Solving for \( \theta \), we find \( \theta = \arctan\left(\frac{-5.52}{8.63}\right) \approx -32.5^\circ \), meaning the force is directed below the \( x \)-axis. This visualization helps predict the resultant impact on the charge.

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Most popular questions from this chapter

A negative point charge \(q_{1}=-4.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=0.60 \mathrm{m} .\) A second point charge \(q_{2}\) is on the \(x\) -axis at \(x=-1.20 \mathrm{m} .\) What must the sign and magnitude of \(q_{2}\) be for the net electric field at the origin to be (a) 50.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -direction and \((\mathrm{b}) 50.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) -x-direction?

CP Two identical spheres are each attached to silk threads of length \(L=0.500 \mathrm{m}\) and hung from a common point (Fig. P21.68). Each sphere has mass \(m=8.00 \mathrm{g} .\) The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_{1},\) and the other a different positive charge \(q_{2} ;\) this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta=20.0^{\circ}\) with the vertical. (a) Draw a free-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the information you have been given, what can you say about the magnitudes of \(q_{1}\) and \(q_{2} ?\) Explain your answers. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of \(30.0^{\circ}\) with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

(a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A\) ? (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s},\) east. What is the speed of the proton at point \(B\) ?

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.

CALC Two thin rods of length \(L\) lie along the \(x\) -axis, one between \(x=a / 2\) and \(x=a / 2+L\) and the other between \(x=-a / 2\) and \(x=-a / 2-L .\) 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 $$F=\frac{Q^{2}}{4 \pi \epsilon_{0} L^{2}} \ln \left[\frac{(a+L)^{2}}{a(a+2 L)}\right]$$ (c) Show that if \(a>>L,\) the magnitude of this force reduces to \(F=Q^{2} / 4 \pi \epsilon_{0} a^{2} .\) (Hint: Use the expansion \(\ln (1+z)=z-\) \(z^{2} / 2+z^{3} / 3-\cdots,\) valid for \(|z|<<1 .\) Carry all expansions to at least order \(L^{2} / a^{2} .\) ) Interpret this result.
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