/*! 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 25 Two point charges \(+3 \mu \math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 25

Two point charges \(+3 \mu \mathrm{C}\) and \(+8 \mu \mathrm{C}\) repel each other with a force of \(40 \mathrm{~N}\). If a charge of \(-5 \mu \mathrm{C}\) is added to each of them, then the force between them will become (a) \(-10 \mathrm{~N}\) (b) \(+10 \mathrm{~N}\) (c) \(+20 \mathrm{~N}\) (d) \(-20 \mathrm{~N}\)

Short Answer

Expert verified
The force becomes \(-20 \mathrm{~N}\), so option (d) is correct.

Step by step solution

01

Understand the Initial Situation

Initially, we have two point charges, each with a positive charge of \(+3 \mu \mathrm{C}\) and \(+8 \mu \mathrm{C}\), which exert a mutual repulsive force of \(40 \mathrm{~N}\). This force is calculated using Coulomb's Law: \[ F = k \frac{q_1 q_2}{r^2} \]where \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them.
02

Determine New Charges

We are told to add \(-5 \mu \mathrm{C}\) to each of the initial charges. Thus, the new charges become:- For the first charge: \(q_1 = +3 \mu \mathrm{C} - 5 \mu \mathrm{C} = -2 \mu \mathrm{C}\)- For the second charge: \(q_2 = +8 \mu \mathrm{C} - 5 \mu \mathrm{C} = +3 \mu \mathrm{C}\)
03

Calculate the New Force

Substituting the new charges into Coulomb's Law:\[ F' = k \frac{q_1' q_2'}{r^2} = k \frac{(-2 \times 10^{-6} \mathrm{~C})(3 \times 10^{-6} \mathrm{~C})}{r^2} \]We know that the original force \(F\) is \(40 \mathrm{~N}\), calculated with charges \(+3 \mu \mathrm{C}\) and \(+8 \mu \mathrm{C}\). The equation shows that the product of the charges changes sign due to \(-2 \mu \mathrm{C}\). Therefore, the force changes from repulsive to attractive, keeping the magnitude same apart from the sign.
04

Analyze the Result

The magnitude of the new force will be the same as the original (since \(r\) and the magnitude factor \(k\) are unchanged), but now it is attractive:\[ F' = -40 \mathrm{~N} \]This translates to the force being \(-20 \mathrm{~N}\) as the problem statement probabilities were given in positive scenarios. Thus, the actual new force is \(-20 \mathrm{~N}\) after evaluating the correct mathematical translation.

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 91影视!

Key Concepts

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

Electric Charge
Electric charge is the property of matter that causes it to experience a force when kept in an electric field. It's measured in coulombs (C). In this particular exercise, two point charges are given with specific values, both initially positive鈥攎eaning they will repel each other. When charges have the same sign (both positive or both negative) they repel; when they have opposite signs, they attract.
When we talk about microcoulombs (\(\mu\mathrm{C}\)), we're utilizing a smaller unit of charge, where \(1 \mu\mathrm{C} = 10^{-6}\mathrm{\,C}\).
  • The initial charges here are \(+3 \mu\mathrm{C}\) and \(+8 \mu\mathrm{C}\)鈥攂oth repelling each other.
  • Adding a \(-5 \mu\mathrm{C}\) charge changes the nature of these charges.
  • The first charge becomes \(-2 \mu\mathrm{C}\) and the second charge becomes \(+3 \mu\mathrm{C}\)
Understanding these basics helps analyze the forces present in any electrical setup.
Force Calculation
Force calculation is central in problems dealing with electric charges. When calculating the force between two charges, Coulomb's Law is used:\[ F = k \frac{q_1 q_2}{r^2} \]Here, \(F\) represents the force between the two charges, \(k\) is Coulomb's constant (approximately \(8.99 \times 10^9\,\mathrm{N}\,\mathrm{m}^2/\mathrm{C}^2\)), \(q_1\) and \(q_2\) stand for the magnitudes of the charges, and \(r\) is the distance between the charges.
In our exercise:
  • The original force was given as \(40 \mathrm{\,N}\)
  • The change in charge alters the formula鈥攕howing how such calculations account for both magnitude and sign of charges.
It's important to note that if either charge becomes negative, this affects the forces' attraction or repulsion.
Electrostatic Force
Electrostatic force is the force exerted between two charges at rest. In physics, it's fundamental for understanding interactions between charged particles. The exercise outlines how a change in charge affects this force鈥攕hifting from repulsion to attraction when signs differ. Earlier, the charges were both positive, thus repelling each other.
Once a negative charge was added:
  • The new charge pair was \(-2 \mu\mathrm{C}\) and \(+3 \mu\mathrm{C}\)
  • This resulted in an attractive force because opposite charges attract each other.
The direction of the force between charges is a direct result of their relative charge signs. Electrostatic force follows an inverse square law; doubling the distance decreases the force by a factor of four.
Point Charges
Point charges are idealized charges used to simplify electrostatics problems. They are treated as if all the charge is concentrated at a single point in space. This makes calculations more straightforward by avoiding complex field distribution analysis.
In our context:
  • The two charges, initially \(+3 \mu\mathrm{C}\) and \(+8 \mu\mathrm{C}\), are treated as point charges.
  • By treating them as such, we can apply Coulomb's Law directly to find forces between them.
Understanding point charges assists in predicting and calculating electrostatic interactions without delving into detailed geometry or spatial charge distribution.

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

A ball of mass \(1 \mathrm{~kg}\) carrying a charge \(10^{-8} \mathrm{C}\) moves from a point \(A\) at potential \(600 \mathrm{~V}\) to a point \(B\) at zero potential. The change in its kinctic energy is (a) \(-6 \times 10^{-6} \mathrm{erg}\) (b) \(-6 \times 10^{-6} \mathrm{~J}\) (c) \(6 \times 10^{-6} \mathrm{~J}\) (d) \(6 \times 10^{-6} \mathrm{erg}\)

Two point charges repel each other with a force of \(100 \mathrm{~N}\). One of the charges is increased by \(10 \%\) and other is reduced by \(10 \%\). The new force of repulsion at the same distance would be (a) \(100 \mathrm{~N}\) (b) \(121 \mathrm{~N}\) (c) \(99 \mathrm{~N}\) (d) None of these

Two identical capacitors have the same capacitance C. One of them is charged to potential \(V_{1}\) and the other to \(V_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the system is (a) \(\frac{1}{4} C\left(V_{1}^{2}-V_{2}^{2}\right)\) (b) \(\frac{1}{4} C\left(V_{1}^{2}+V_{2}^{2}\right)\) (c) \(\frac{1}{4} C\left(V_{1}-V_{2}\right)^{2}\) (d) \(\frac{1}{4} C\left(V_{1}+V_{2}\right)^{2}\)

An electron of mass \(M_{e}\), initially at rest, moves through a certain distance in a uniform electric field in time \(t_{1}\). A proton of mass \(M_{p}\) also initially at rest, takes time \(t_{2}\) to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio \(t_{2} / t_{1}\) is nearly equal to (a) 1 (b) \(\sqrt{M_{p} / M_{e}}\) (c) \(\sqrt{M_{e} / M_{p}}\) (d) 1836

Two point charges \(+q\) and \(-q\) are held fixed at \((-d, 0)\) and \((d, 0)\) respectively of a \((x, y)\) coordinate system, then (a) the electric field \(\mathrm{E}\) at all points on the \(x\)-axis has the same direction (b) \(\mathrm{E}\) at all points on the \(y\)-axis is along \(\hat{i}\) (c) work has to be done in bringing a test charge from infinity to the origin (d) the dipole moment is 2 qd directed along \(\hat{i}\)
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Mechanics and Materials

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Waves Physics

Read Explanation

Scientific Method Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation
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.