Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 84

Two metal spheres of radius \(5.0 \mathrm{cm}\) carry net charges of $+1.0 \mu \mathrm{C}\( and \)+0.2 \mu \mathrm{C} .$ (a) What (approximately) is the magnitude of the electrical repulsion on either sphere when their centers are \(1.00 \mathrm{m}\) apart? (b) Why cannot Coulomb's law be used to find the force of repulsion when their centers are \(12 \mathrm{cm}\) apart? (c) Would the actual force be larger or smaller than the result of using Coulomb's law with \(r=12 \mathrm{cm} ?\) Explain.

Short Answer

Expert verified
Based on the provided information: 1. The electrical repulsion between the two spheres is approximately 1.80 N when their centers are 1.00 m apart. 2. Coulomb's Law cannot be used to find the force of repulsion when the distance between their centers is 12 cm because it assumes point charges, and the interacting charge distribution on the spheres' surfaces must be considered at this distance. 3. When the distance between the centers of the spheres is 12 cm, the actual force is *smaller* than the result using Coulomb's Law because the charges redistribute themselves on the spheres' surfaces, leading to a weaker repulsive force.

Step by step solution

01

Finding the electrical repulsion at 1.00 m

We can use Coulomb's Law to find the electrical repulsion between two charged objects at a certain distance. Coulomb's Law is given by: $$ F = k \frac{q_1q_2}{r^2} $$ Where: - \(F\) is the force (magnitude of the electrical repulsion), - \(k\) is the Coulomb constant (\(=8.9875 x 10^9 Nm^2C^{-2}\)), - \(q_1\) and \(q_2\) are the charges of the two objects, - \(r\) is the distance between the centers of the two objects. In this case: - \(q_1 = 1.0 \mu C\) - \(q_2 = 0.2 \mu C\) - \(r = 1.00 m\) We can now plug in the values and compute the force.
02

Computing the magnitude of the electrical repulsion at 1.00 m

Substituting the given values into Coulomb's Law, we get: $$ F = (8.9875 x 10^9 Nm^2C^{-2}) \frac{(1.0 \times 10^{-6} C)(0.2 \times 10^{-6} C)}{(1.00 m)^2} $$ By evaluating this expression, we find the magnitude of the electrical repulsion: $$ F \approx 1.80 N $$ So, the electrical repulsion between the two spheres is approximately \(1.80\) N when their centers are \(1.00\) m apart.
03

Explaining why Coulomb's Law cannot be used at 12 cm distance

Coulomb's Law assumes that charged objects are point charges, which means that their charges are concentrated at a single point. When the spheres are very close to each other (in this case, when their centers are only \(12\) cm apart), the charges on the spheres' surfaces interact with each other, and it becomes necessary to consider the charge distribution over the entire surface of the spheres. This interacting charge distribution cannot be ignored and results in a more complex force calculation. Therefore, Coulomb's Law is not applicable when the centers of the spheres are \(12\) cm apart.
04

Determining the actual force at 12 cm distance

When the centers of the spheres are \(12\) cm apart, their actual charges are distributed over their surface. Since both spheres are charged positively, their nearby charges will repel each other and the charges will redistribute themselves on the spheres' surfaces. This redistribution causes charges on the spheres' surfaces closer to each other to be lower than the average, leading to a weaker repulsive force than the Coulomb's Law approximation. Therefore, the actual force at a \(12\) cm distance is *smaller* than the result using Coulomb's Law at that distance.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Some forms of cancer can be treated using proton therapy in which proton beams are accelerated to high energies, then directed to collide into a tumor, killing the malignant cells. Suppose a proton accelerator is \(4.0 \mathrm{m}\) long and must accelerate protons from rest to a speed of $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} .$ Ignore any relativistic effects (Chapter 26) and determine the magnitude of the average electric field that could accelerate these protons.
A metallic sphere has a charge of \(+4.0 \mathrm{nC} .\) A negatively charged rod has a charge of \(-6.0 \mathrm{nC} .\) When the rod touches the sphere, $8.2 \times 10^{9}$ electrons are transferred. What are the charges of the sphere and the rod now?
A balloon, initially neutral, is rubbed with fur until it acquires a net charge of \(-0.60 \mathrm{nC} .\) (a) Assuming that only electrons are transferred, were electrons removed from the balloon or added to it? (b) How many electrons were transferred?
A hollow conducting sphere of radius \(R\) carries a negative charge \(-q .\) (a) Write expressions for the electric field \(\overrightarrow{\mathbf{E}}\) inside \((rR)\) the sphere. Also indicate the direction of the field. (b) Sketch a graph of the field strength as a function of \(r\). [Hint: See Conceptual Example \(16.8 .]\)
An electron traveling horizontally from west to east enters a region where a uniform electric field is directed upward. What is the direction of the electric force exerted on the electron once it has entered the field?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Wave Optics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Magnetism

Read Explanation

Geometrical and Physical Optics

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.

Sign-up for free