/*! 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 31 A uniform line of charge with le... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 31

A uniform line of charge with length \(20.0 \mathrm{~cm}\) is along the \(x\) -axis, with its midpoint at \(x=0 .\) Its charge per length is \(+4.80 \mathrm{nC} / \mathrm{m}\) A small sphere with charge \(-2.00 \mu \mathrm{C}\) is located at \(x=0, y=5.00 \mathrm{~cm}\) What are the magnitude and direction of the force that the charged sphere exerts on the line of charge?

Short Answer

Expert verified
To obtain the exact value of the force and its direction one needs to perform the above integrations numerically. Hence, without exact values, the answer can be stated as follows: The direction of the force is opposite to the electric field and the magnitude of the force is equal to the product of the total electric field and the sphere's charge.

Step by step solution

01

Determine the electric field due to a small charge element

Let's consider a small element of the line charge with length \(dx\) at a distance \(x\) from the sphere. The charge \(dq\) of this element can be calculated as \(dq=\lambda dx\), where \(\lambda=+4.80 \, nC/m\) is the charge per length. The electric field \(dE\) due to \(dq\) at the location of the sphere is \(dE=k \frac{dq}{r^2}\), where \(k = 8.99脳10^9 N m^2/C^2\) is the Coulomb's constant and \(r = \sqrt{x^2+(5.00 cm)^2}\) is the distance between \(dq\) and the sphere.
02

Integrate the electric field over the whole line of charge

Because the electric field is a vector quantity, we should separate it into its y and x components and then integrate them separately over the length of the line charge. Mathematically, this integration can be expressed as \(E_y = \int -dE cos\theta\), \(E_x = \int -dE sin\theta\), where \(cos\theta = x/r\), \(sin\theta = 5.00 cm/r\). The limits of integration would be from \(-10 cm\) to \(10 cm\), as the line charge spans \(20 cm\) centered on the x-axis.
03

Calculate the force exerted by the sphere on the line of charge

The force that the charged sphere exerts on the line of charge can be calculated as \(F = qE\), where \(q = -2.00 \mu C\) is the charge of the sphere, and \(E\) is the total electric field which is \(\sqrt{{E_x}^2+{E_y}^2}\).
04

Determine the direction of the force

The direction of the force can be calculated using trigonometry. The angle 蠁 between the direction of the force and the x-axis can be found as \(蠁 = atan(E_y/E_x)\). The direction of the force will be the opposite to the direction of the electric field, because the sphere's charge is negative.

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.

Line Charge
A line charge is a distribution of electric charge along a line in space. In our problem, this charge is uniformly spread along a line on the x-axis, creating an electric field around it.
A line charge can be imagined as a thin wire carrying a continuous charge along its length. The charge per unit length is called linear charge density, denoted by the symbol \(\lambda\).
For our exercise:
  • The line's total length is 20.0 cm, positioned symmetrically about the origin (x = 0).
  • The linear charge density \(\lambda\) is given as \(+4.80 \, \text{nC/m}\).
Understanding the concept of line charge is vital for calculating the electric field it produces. This field interacts with other charges, like the small sphere mentioned in the problem, exerting forces upon them.
Coulomb's Law
Coulomb's Law describes how the electric force acts between charged objects. It's a foundational principle that helps us understand how charges influence each other.
Mathematically, Coulomb's Law is given by:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
where \( F \) is the force between the charges, \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, N \, m^2/C^2 \), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.
In our problem, this law helps calculate the electric field \( dE \) due to each small charge element \( dq \) existing on the line. By summing (or integrating) these small fields from each charge segment, we find the net electric field acting on the charged sphere. Then, using this electric field, we can find the force the sphere's charge exerts back on the line of charge.
Vector Components
Understanding vector components is crucial when dealing with electric fields and forces. Electric fields are vector quantities, meaning they have both magnitude and direction.
To solve the given problem, we break the electric field into x and y components. This separation is essential because the field's effect depends upon its direction relative to the charge.
  • The x-component \( E_x \) of the electric field is found using the sine component: \( E_x = - \int dE \sin \theta \).
  • The y-component \( E_y \) of the electric field is calculated using the cosine component: \( E_y = - \int dE \cos \theta \).
The angle \( \theta \) relates to the position of a differential charge element on the line, calculated using trigonometry. By integrating these components over the entire line charge, we produce the complete electric field. This result allows us to find both the magnitude and direction (via the arctangent function) of the force on the sphere, accounting for its negative charge.

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

Point charge \(A\) is on the \(x\) -axis at \(x=-3.00 \mathrm{~cm}\). At \(x=1.00 \mathrm{~cm}\) on the \(x\) -axis its electric field is \(2700 \mathrm{~N} / \mathrm{C}\). Point charge \(B\) is also on the \(x\) -axis, at \(x=5.00 \mathrm{~cm}\). The absolute magnitude of charge \(B\) is twice that of \(A .\) Find the magnitude and direction of the total electric field at the origin if (a) both \(A\) and \(B\) are positive; (b) both are negative; (c) \(A\) is positive and \(B\) is negative; (d) \(A\) is negative and \(B\) is positive.

Neurons are components of the nervous system of the body that transmit signals as electrical impulses travel along their length. These impulses propagate when charge suddenly rushes into and then out of a part of the neuron called an axon. Measurements have shown that, during the inflow part of this cycle, approximately \(5.6 \times 10^{11} \mathrm{Na}^{+}\) (sodium ions) per meter, each with charge \(+e,\) enter the axon. How many coulombs of charge enter a \(1.5 \mathrm{~cm}\) length of the axon during this process?

A proton is placed in a uniform electric field of \(2.75 \times 10^{3} \mathrm{~N} / \mathrm{C} .\) Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after \(1.00 \mu \mathrm{s}\) in the field, assuming it starts from rest.

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of the field point (a) at \(x=0, y=-1.35 \mathrm{~m}\) (b) at \(x=12.0 \mathrm{~cm}, y=12.0 \mathrm{~cm} ;\) (c) at \(x=-1.10 \mathrm{~m}, y=2.60 \mathrm{~m} ?\) Express your results in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\)

A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{~m} / \mathrm{s}\). (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of \(3.20 \mathrm{~cm}\). (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Circular Motion and Gravitation

Read Explanation

Dynamics

Read Explanation

Solid State Physics

Read Explanation

Electricity

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.