
Concept explainers
Example 23.3 derives the exact expression for the electric field at a point on the axis of a uniformly charged disk. Consider a disk of radius R = 3.00 cm having a uniformly distributed charge of +5.20 μC. (a) Using the result of Example 23.3, compute the electric field at a point on the axis and 3.00 mm from the center. (b) What If? Explain how the answer to part (a) compares with the field computed from the near-field approximation E = σ/2ϵ0. (We derived this expression in Example 23.3.) (c) Using the result of Example 23.3, compute the electric field at a point on the axis and 30.0 cm from the center of the disk. (d) What If? Explain how the answer to part (c) compares with the electric field obtained by treating the disk as a +5.20-μC charged particle at a distance of 30.0 cm.
(a)

The electric field at a point on the axis at a distance
Answer to Problem 5P
The electric field at a point on the axis and
Explanation of Solution
The value of the charge is
The formula to calculate the surface charge density is,
The formula to calculate the area is,
Substitute
Substitute
The formula to calculate the electric field at a point from the center of a uniformly charged disk is,
Here,
Conclusion:
Substitute
Therefore, the electric field at a point on the axis and
(b)

The change in electric field when calculated using near field approximation.
Answer to Problem 5P
The magnitude of electric field when computed from near field approximation is
Explanation of Solution
The formula to calculate the electric field at a point from the center of a uniformly charged disk is,
Here,
Conclusion:
Substitute
The value of electric field for uniformly charged disk for a point
The formula to calculate the percentage change with respect to the field computed from near field approximation is,
Therefore, the magnitude of electric field when computed from near field approximation is
(c)

The electric field at a point on the axis at a distance
Answer to Problem 5P
The electric field at a point on the axis and
Explanation of Solution
The value of the charge is
The formula to calculate the surface charge density is,
The formula to calculate the area is,
Substitute
Substitute
The formula to calculate the electric field at a point from the center of a uniformly charged disk is,
Here,
Conclusion:
Substitute
Therefore, the electric field at a point on the axis and
(d)

The change in electric field at
Answer to Problem 5P
The magnitude of electric field obtained by treating the disk as a
Explanation of Solution
The formula to calculate the electric field of a charged particle is,
Here,
Substitute
the percentage change when part (c) is compared with the field obtained by treating the disk as a
The value of electric field for uniformly charged disk at a point
The value of the electric field while approximating the disc to be a point charge is
Conclusion:
The formula to calculate the percentage change when part (c) is compared with the field obtained by treating the disk as a
Therefore, the magnitude of electric field obtained by treating the disk as a point charge at a distance of
Want to see more full solutions like this?
Chapter 23 Solutions
PHYSICS:F/SCI.+ENGRS.(LL)-W/WEBASSIGN
- No chatgpt pls will upvotearrow_forwardYou are standing a distance x = 1.75 m away from this mirror. The object you are looking at is y = 0.29 m from the mirror. The angle of incidence is θ = 30°. What is the exact distance from you to the image?arrow_forwardFor each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank youarrow_forward
- A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).arrow_forwardWhat are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V) ammeter I =arrow_forwardsimple diagram to illustrate the setup for each law- coulombs law and biot savart lawarrow_forward
- A circular coil with 100 turns and a radius of 0.05 m is placed in a magnetic field that changes at auniform rate from 0.2 T to 0.8 T in 0.1 seconds. The plane of the coil is perpendicular to the field.• Calculate the induced electric field in the coil.• Calculate the current density in the coil given its conductivity σ.arrow_forwardAn L-C circuit has an inductance of 0.410 H and a capacitance of 0.250 nF . During the current oscillations, the maximum current in the inductor is 1.80 A . What is the maximum energy Emax stored in the capacitor at any time during the current oscillations? How many times per second does the capacitor contain the amount of energy found in part A? Please show all steps.arrow_forwardA long, straight wire carries a current of 10 A along what we’ll define to the be x-axis. A square loopin the x-y plane with side length 0.1 m is placed near the wire such that its closest side is parallel tothe wire and 0.05 m away.• Calculate the magnetic flux through the loop using Ampere’s law.arrow_forward
- Describe the motion of a charged particle entering a uniform magnetic field at an angle to the fieldlines. Include a diagram showing the velocity vector, magnetic field lines, and the path of the particle.arrow_forwardDiscuss the differences between the Biot-Savart law and Coulomb’s law in terms of their applicationsand the physical quantities they describe.arrow_forwardExplain why Ampere’s law can be used to find the magnetic field inside a solenoid but not outside.arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning





