/*! 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 89 Two small spheres, each carrying... [FREE SOLUTION] | 91Ó°ÊÓ

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Two small spheres, each carrying a net positive charge, are separated by \(0.400 \mathrm{~m}\). You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (charge \(q_{1}\) ) at the origin and the other sphere (charge \(q_{2}\) ) at \(x=+0.400 \mathrm{~m}\). Available to you are a third sphere with net charge \(q_{3}=4.00 \times 10^{-6} \mathrm{C}\) and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the \(x\) -axis at \(x=0.200 \mathrm{~m} ;\) you measure the net force on it to be \(4.50 \mathrm{~N}\) in the \(+x\) -direction. Then you move the third sphere to \(x=+0.600 \mathrm{~m}\) and measure the net force on it now to be \(3.50 \mathrm{~N}\) in the \(+x\) -direction. (a) Calculate \(q_{1}\) and \(q_{2}\). (b) What is the net force (magnitude and direction) on \(q_{3}\) if it is placed on the \(x\) -axis at \(x=-0.200 \mathrm{~m} ?\) (c) At what value of \(x\) (other than \(x=\pm \infty\) ) could \(q_{3}\) be placed so that the net force on it is zero?

Short Answer

Expert verified
The charges on spheres 1 and 2 correspond to the solution of the system of equations obtained in Step 1. The force on sphere 3 at the new location equals to the solution from Step 2. The position \( x \) at which the force on sphere 3 is zero corresponds to the solution of the equation obtained in Step 3.

Step by step solution

01

Determine Charges of Sphere 1 and 2

At the first position we get the equation \( F = q_3 . q_1/r_1^2 = k . q_3 . q_2 / r_2^2 \), where \( r_1 = 0.2m, r_2 = 0.2m \) and \( F = 4.5N \). Solving this gives us an equation in \( q_1 \) and \( q_2 \). At the second position, we get similar equation with \( r_1 = 0.6m \) and \( r_2 = 0.2m \). Solving these two equations together would give us the values of \( q_1 \) and \( q_2 \).
02

Determine Force on Sphere 3 at New Position

We use the calculated charges and Coulomb's law again to calculate force at \( x = -0.2m \). The distances for sphere 1 and sphere 2 now are 0.2m and 0.6m respectively.
03

Determine Position at Which Force on Sphere 3 is Zero

We solve for the position \( x \) which equates the forces due to Sphere 1 and Sphere 2. As Force \( F = k.q_1.q_3/x^2 = k.q_2.q_3/(0.4-x)^2 \), we can equate the two forces and solve for \( x \).

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Key Concepts

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

Electric Charge
Electric charges are a fundamental property of matter responsible for electric force and electromagnetic interactions. Every charged particle has an electric charge, measured in Coulombs (C). There are two types of electric charges: positive and negative. Like charges repel each other, while opposite charges attract each other. In this problem, all spheres have a positive charge, which is why they repel each other.
A charged sphere can have a net positive charge if it has more protons than electrons, leading to a net positive charge like the spheres in this exercise. Understanding electric charge is crucial because it plays a significant role in determining the electric forces between objects.
Electric Force
Electric force is the force of attraction or repulsion between two charged objects. It is a part of the four fundamental forces of nature and is governed by Coulomb's Law. In the exercise, the spheres experience electric force due to their charges positioned along the x-axis.
Coulomb's Law describes this force with the formula: \[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]where \( F \) is the electric force, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \) N\( \cdot \)m²/C²), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
In this scenario, the force was measured on sphere \( q_3 \) using its distance from \( q_1 \) and \( q_2 \), and the values of these forces help us determine the unknown charges through careful application of this principle.
Electrostatics
Electrostatics is the branch of physics that studies electric charges at rest and the forces between them. It is crucial for understanding scenarios like those described in this problem, where charges exert forces on one another at a specific distance. In this problem, electrostatics principles help calculate unknown charges and predict the net force on the third sphere.
In analyzing electrostatics problems such as this, the principle of superposition is often used. It states that the net force on any charge is the vector sum of forces exerted by other individual charges. This principle allows us to calculate the net force on \( q_3 \) when placed at different positions by summing up the individual forces due to \( q_1 \) and \( q_2 \).
Electrostatic scenarios assume that the charges remain fixed in position and do not move under the influence of forces, making it possible to solve the problem with static conditions.
Physics Problem Solving
Physics problem solving often involves breaking down problems into simpler parts and using systematic methods to find solutions. In this exercise, the application of Coulomb's Law provides a framework for calculating unknown charges.
Effective problem-solving strategies in physics include:
  • Identifying knowns and unknowns: Distinguishing between what is given (force, distances, charge of sphere \( q_3 \)) and what needs to be calculated (\( q_1 \) and \( q_2 \)).
  • Applying known principles: Utilizing the principles of Coulomb’s Law and superposition for calculations.
  • Setting up equations systematically: Deriving equations based on force measurements taken at different positions.
  • Solving step-by-step: Working through the mathematical manipulations required to find the final answer.
These strategies help create a clear pathway to solving complex physics problems, turning seemingly challenging questions into manageable tasks.

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Most popular questions from this chapter

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by \(3.1 \mathrm{~mm}\), 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 \(7.2 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m} ?\)

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. Ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s},\) and \(\alpha=30.0^{\circ}\)

A small sphere with mass \(m\) carries a positive charge \(q\) and is attached to one end of a silk fiber of length \(L .\) The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density \(\sigma\). Show that when the sphere is in equilibrium, the fiber makes an angle equal to arctan \(\left(q \sigma / 2 m g \epsilon_{0}\right)\) with the vertical sheet.

A very long line of charge with charge per unit length \(+8.00 \mu \mathrm{C} / \mathrm{m}\) is on the \(x\) -axis and its midpoint is at \(x=0 .\) A second very long line of charge with charge per length \(-4.00 \mu \mathrm{C} / \mathrm{m}\) is parallel to the \(x\) -axis at \(y=10.0 \mathrm{~cm}\) and its midpoint is also at \(x=0 .\) At what point on the \(y\) -axis is the resultant electric field of the two lines of charge equal to zero?

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(1.60 \mathrm{~cm}\) distant from the first, in a time interval of \(3.20 \times 10^{-6} \mathrm{~s} .\) (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

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