Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

$+ Q$ In Fig. 22-30a, a circular plastic rod with uniform charge $+ Q$ produces an electric field of magnitude E at the center of curvature (at the origin). In Figs. 22-30b, c, and d, more circular rods, each with identical uniform charges , are added until the circle is complete. A fifth arrangement (which would be labeled e) is like that in d except the rod in the fourth quadrant has charge $- Q$ . Rank the five arrangements according to the magnitude of the electric field at the center of curvature, greatest first.

The rank of the five arrangements according to the magnitude of the electric field at the center of curvature is $E_{5} > E_{2} > E_{1} = E_{3} > E_{4}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Understanding the concept of wave

The electric field at the center of curvature in a given quadrant depends on the half-angle of the quadrant. Using this data, the electric field in all three cases is calculated and compared to get the required rank value.

The total electric field at the center of curvature of a circular plastic rod,

$\begin{array}{l} E = \frac{2 k λ}{r} \sin \frac{θ}{2} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (1) \\ λ = line charge density, \\ r = distance of point from the curve, \\ θ = the angle of quadrant \end{array}$

Step 2: Calculation of the rank according to electric field magnitudes at the center of curvature

Now, for case a, the magnitude of the electric field at the center can be given using equation (i) as $(θ = \frac{π}{2})$ :

$\begin{matrix} E_{1} = \frac{2 k λ}{r} \sin \frac{π}{4} \\ = \frac{2 k λ}{r} (\frac{1}{\sqrt{2}}) \\ = \frac{\sqrt{2} k λ}{r} \end{matrix}$

Now, for case b, the magnitude of the electric field at the center can be given using equation (i) as $(θ = π)$ :

$\begin{matrix} E_{2} = \frac{2 k λ}{r} \sin \frac{π}{2} \\ = \frac{2 k λ}{r} \end{matrix}$

Now, for case c, the magnitude of the electric field at the center can be given using equation (i) as $(θ = \frac{3 π}{2})$ :

$\begin{matrix} | E_{3} | = | \frac{2 k λ}{r} \sin \frac{3 π}{4} | \\ = \frac{2 k λ}{r} (\frac{1}{\sqrt{2}}) \\ = \frac{\sqrt{2} k λ}{r} \end{matrix}$

Now, for case d, the magnitude of the electric field at the center can be given using equation (i) as: $(θ = 2 π)$

$\begin{matrix} E_{4} = \frac{2 k λ}{r} \sin π \\ = 0 \end{matrix}$

Now, for the additional fifth arrangement where the fourth quadrant has charge, the magnitude of electric field using equation (i) can be written as: (due to a negative quadrant we can get the net field as the sum of three case that is field due to three quadrant, field due to negative quadrant, and field due to the additional quadrant)

$\begin{matrix} E_{5} = | \frac{2 k λ}{r} \sin \frac{3 π}{4} | - \frac{2 k λ}{r} \sin \frac{π}{2} + \frac{2 k λ}{r} \sin \frac{π}{4} \\ = \frac{4 k λ}{r} (\frac{1}{\sqrt{2}}) \\ = \frac{2 \sqrt{2} k λ}{r} \end{matrix}$

Hence, the rank of the arrangements according to the magnitudes of electric field is $E_{5} > E_{2} > E_{1} = E_{3} > E_{4}$ .

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Understanding the concept of wave

Step 2: Calculation of the rank according to electric field magnitudes at the center of curvature

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Understanding the concept of wave

Step 2: Calculation of the rank according to electric field magnitudes at the center of curvature

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Famous Physicists

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics