Question
Book Icon
Chapter 33, Problem 1P

(a)

To determine

The time rate of increase of electric field between the plates.

(a)

Expert Solution
Check Mark

Answer to Problem 1P

The time rate of increase of electric field between the plates is 7.19×1011V/ms .

Explanation of Solution

Given info: The current is 0.200A , the radius of the circular plate is 10.0cm and the separation distance between the plates is 4.00mm .

The value of permittivity of free space is 8.85×1012F/m and the value of the permeability constant is ×107H/m

The area of the circular plate is,

A=πr2

Here,

r is the radius of the circular plate.

The formula for the capacitance of the capacitor is.

C=Aε0d=πr2ε0d

Here,

ε0 is the permittivity of free space.

d is the separation between the plates.

Substitute 10.0cm for r , 8.85×1012F/m for ε0 and 4.00mm for d in above equation to find C .

C=π(10.0cm(102m1cm))2(8.85×1012F/m)4.00mm(103m1mm)=6.95×1011F

Thus, the capacitance of the capacitor is 6.95×1011F .

The formula for the capacitance of the capacitor is.

C=QVV=QC

Here,

Q is the charge of the plates.

V is the potential of the plates.

Differentiate the above equation with respect to time.

dVdt=dQdtC

The formula for the current is,

I=dQdt

Then,

dVdt=IC

The formula for the electric field is,

E=Vd

Differentiate the above equation with respect to time.

dEdt=1ddVdt

Substitute IC for dVdt in above equation.

dEdt=ICd

Substitute 0.200A for I , 6.95×1011F for C and 4.00mm for d in above equation to find dEdt .

dEdt=0.200A(6.95×1011F)(4.00mm(103m1mm))=7.19×1011V/ms

Conclusion:

Therefore, the time rate of increase of electric field between the plates is 7.19×1011V/ms .

(b)

To determine

The magnetic field between the plates 5.00cm from the centre.

(b)

Expert Solution
Check Mark

Answer to Problem 1P

The magnetic field between the plates 5.00cm from the centre is 2.0×107T .

Explanation of Solution

Given info: The current is 0.200A , the radius of the circular plate is 10.0cm and the separation distance between the plates is 4.00mm .

The formula for the electric flux through the circular plate is,

ΦE=EAdΦEdt=AdEdt=πr2dEdt

The Maxwell’s equation is.

Bds=ε0μ0dΦEdtB(2πr)=ε0μ0πr2dEdtB=(rε0μ02)dEdt

Substitute 5.00cm for r , 8.85×1012F/m for ε0 , ×107H/m for μ0 and 7.19×1011V/ms for dEdt in above equation to find B .

B=((5.00cm(102m1cm))(8.85×1012F/m)(×107H/m)2)(7.19×1011V/ms)=1.9990×107T2.0×107T

Conclusion:

Therefore, the magnetic field between the plates 5.00cm from the centre is 2.0×107T .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC
In the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?
Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor  µC 6.00 µF capacitor  µC 3.00 µF capacitor  µC capacitor C  µC

Chapter 33 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Multi-term

Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning