Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Question
Chapter 24, Problem 65CP
To determine
The electric field within the charge distribution in the spherical shell.
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A cylinder of length L=5m has a radius R=2 cm and linear charge density 2=300 µC/m. Although the
linear charge density is a constant through the cylinder, the charge density within the cylinder changes
with r. Within the cylinder, the charge density of the cylinder varies with radius as a function p( r) =p.r/R.
Here R is the radius of the cylinder and R=2 cm and p, is just a constant that you need to determine.
b. Find the constant po in terms of R and 2. Then plug in values of R and 1. to find the value for
the constant p.
c. Assuming that L>>R, use Gauss's law to find out the electric field E inside the cylinder (rR) in terms of 1. and R.
d. Based on your result from problem c, find the electric field E at r=1cm and r=4cm.
Consider any charge distribution with a charge density e(7), let v be a
spherical region of radius Y.. Centered at 0, the average electric field E
within v has the formula:
1
E:
Iar / E(F)dv = Ent + Eot
%3D
where Ent is the average field due to all internal charges v and Eetis the
average field due to all external charges v Prove that
Eint
1
4T€0 r3
Eext
p(T) du'
Jv
4T€0
p13
where p is the electric dipole moment (cycle O) of the internal charge.
and V is the region in space. that includes v and P(F)# 0
3
A non-uniformly charged insulating sphere has a volume charge density p that is expressed as
p= Br
where B is a constant, and r is the radius from the center of the sphere. If the, the total charge of the sphere is Q and its maximum radius is R. What is the value for B?
Sol.
By definition, the volume charge density is expressed infinitesimally as
p=
where in
is the infinitesimal charge and
is the infinitesimal volume.
So, we haye
p = dq/
So we can write this as
dq = B
dV
But,
dV =
dr
By substitution, we get the following
dq = 4BT
dr
Using Integration operation and evaluating its limits, the equation, leads to
Q =
BT
Rearranging, we get
B =
/( T
Chapter 24 Solutions
Physics for Scientists and Engineers with Modern Physics, Technology Update
Ch. 24.1 - Suppose a point charge is located at the center of...Ch. 24.2 - If the net flux through a gaussian surface is...Ch. 24 - Prob. 1OQCh. 24 - Prob. 2OQCh. 24 - Prob. 3OQCh. 24 - Prob. 4OQCh. 24 - Prob. 5OQCh. 24 - Prob. 6OQCh. 24 - Prob. 7OQCh. 24 - Prob. 8OQ
Ch. 24 - Prob. 9OQCh. 24 - Prob. 10OQCh. 24 - Prob. 11OQCh. 24 - Prob. 1CQCh. 24 - Prob. 2CQCh. 24 - Prob. 3CQCh. 24 - Prob. 4CQCh. 24 - Prob. 5CQCh. 24 - Prob. 6CQCh. 24 - Prob. 7CQCh. 24 - Prob. 8CQCh. 24 - Prob. 9CQCh. 24 - Prob. 10CQCh. 24 - Prob. 11CQCh. 24 - A flat surface of area 3.20 m2 is rotated in a...Ch. 24 - A vertical electric field of magnitude 2.00 104...Ch. 24 - Prob. 3PCh. 24 - Prob. 4PCh. 24 - Prob. 5PCh. 24 - A nonuniform electric field is given by the...Ch. 24 - An uncharged, nonconducting, hollow sphere of...Ch. 24 - Prob. 8PCh. 24 - Prob. 9PCh. 24 - Prob. 10PCh. 24 - Prob. 11PCh. 24 - A charge of 170 C is at the center of a cube of...Ch. 24 - Prob. 13PCh. 24 - A particle with charge of 12.0 C is placed at the...Ch. 24 - Prob. 15PCh. 24 - Prob. 16PCh. 24 - Prob. 17PCh. 24 - Find the net electric flux through (a) the closed...Ch. 24 - Prob. 19PCh. 24 - Prob. 20PCh. 24 - Prob. 21PCh. 24 - Prob. 22PCh. 24 - Prob. 23PCh. 24 - Prob. 24PCh. 24 - Prob. 25PCh. 24 - Determine the magnitude of the electric field at...Ch. 24 - A large, flat, horizontal sheet of charge has a...Ch. 24 - Prob. 28PCh. 24 - Prob. 29PCh. 24 - A nonconducting wall carries charge with a uniform...Ch. 24 - A uniformly charged, straight filament 7.00 m in...Ch. 24 - Prob. 32PCh. 24 - Consider a long, cylindrical charge distribution...Ch. 24 - A cylindrical shell of radius 7.00 cm and length...Ch. 24 - A solid sphere of radius 40.0 cm has a total...Ch. 24 - Prob. 36PCh. 24 - Prob. 37PCh. 24 - Why is the following situation impossible? A solid...Ch. 24 - A solid metallic sphere of radius a carries total...Ch. 24 - Prob. 40PCh. 24 - A very large, thin, flat plate of aluminum of area...Ch. 24 - Prob. 42PCh. 24 - Prob. 43PCh. 24 - Prob. 44PCh. 24 - A long, straight wire is surrounded by a hollow...Ch. 24 - Prob. 46PCh. 24 - Prob. 47PCh. 24 - Prob. 48APCh. 24 - Prob. 49APCh. 24 - Prob. 50APCh. 24 - Prob. 51APCh. 24 - Prob. 52APCh. 24 - Prob. 53APCh. 24 - Prob. 54APCh. 24 - Prob. 55APCh. 24 - Prob. 56APCh. 24 - Prob. 57APCh. 24 - An insulating solid sphere of radius a has a...Ch. 24 - Prob. 59APCh. 24 - Prob. 60APCh. 24 - Prob. 61CPCh. 24 - Prob. 62CPCh. 24 - Prob. 63CPCh. 24 - Prob. 64CPCh. 24 - Prob. 65CPCh. 24 - A solid insulating sphere of radius R has a...Ch. 24 - Prob. 67CPCh. 24 - Prob. 68CPCh. 24 - Prob. 69CP
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- A non-uniformly charged insulating sphere has a volume charge density p that is expressed as p= Br where B is a constant, and r is the radius from the center of the sphere. If the, the total charge of the sphere is Q and its maximum radius is R. What is the value for B? Sol. By definition, the volume charge density is expressed infinitesimally as p= where in is the infinitesimal charge and is the infinitesimal volume. so, we have P = dq/ = B so we can write this as dq = B dV But, dV = dr By substitution, we get the following dq = 4B dr Using Integration operation and evaluating its limits, the equation, leads to Q = Rearranging, we get B = 4)arrow_forwardGiven that D = zpcos?o a, C/m², calculate the charge density at (1, 7/4, 3) and the total charge enclosed by the cylinder of radius 1 m with z from -2 m to 2 m.arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rp has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = aer/ao + B/r + bo %3D where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Vab = | S"Edr= - [ *Edr Calculating the antiderivative or indefinite integral, Vab = (-aage-r/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (e-rb/ao - eralao) + B In( ) + bo ( ))arrow_forward
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