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Solutions for Physics, Volume 2, 12th Edition Cutnell (Chapters 18 to 32 included) $29.49   Add to cart

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Solutions for Physics, Volume 2, 12th Edition Cutnell (Chapters 18 to 32 included)

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  • Course
  • Physics - General Relativity
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  • Physics - General Relativity

Complete Solutions Manual for Physics, Volume 1, 12th Edition by John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler ; ISBN13: 9781119803775. Full Chapters included Chapter 18 to 32. Practice Problems included. 18.Electric Forces and Electric Fields. 19.Electric Potential Energy and t...

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  • December 4, 2023
  • 992
  • 2021/2022
  • Exam (elaborations)
  • Questions & answers
  • Physics - General Relativity
  • Physics - General Relativity
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mizhouubcca
Physics
12th Edition
by John D. Cutnell


Volumn - 2

Vol 2 Complete Chapter Solutions
Manual are included (Ch 18 to 32)


** Immediate Download
** Swift Response
** All Chapters included
** Practice Problems

,CHAPTER 18 ELECTRIC FORCES AND
ELECTRIC FIELDS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS


1. 1.9 × 1013

2. (b) Suppose that A is positive and B is negative. Since C and A also attract each other,
C must be negative. Thus, B and C repel each other, because they have like charges (both
negative). Suppose, however, that A is negative and B is positive. Since C and A also
attract each other, C must be positive. Again we conclude that B and C repeal each
other, because they have like charges (both positive).

3. (d) The fact that the positive rod repels one object indicates that that object carries a net
positive charge. The fact that the rod repels the other object indicates that that object
carries a net negative charge. Since both objects are identical and made from conducting
material, they share the combined net charges equally after they are touched together.
Since the rod repels each object after they are touched, each object must then carry a net
positive charge. But the net electric charge of any isolated system is conserved, so the
total net charge initially must also have been positive. This means that the initial positive
charge had the greater magnitude.

4. (c) This distribution is not possible because of the law of conservation of electric charge.
The total charge on the three objects here is 98 q , whereas only q was present initially.

q1 q2
5. (e) Coulomb’s law states that the magnitude of the force is given by F = k . The
r2
force is directed along the line between the charges and is an attraction for unlike charges
and a repulsion for like charges. Charge B is attracted by charge A with a force of
q q
magnitude k 2 and repelled by charge C with a force of the same magnitude. Since
d
q q
both forces point to the left, the net force acting on B has a magnitude of 2k 2 .
d
q q
Charge A is attracted by charge B with a force of k 2 and also by charge C with a
d
q q
force of k 2
. Since both forces point to the right, the net force acting on A has a
( 2d )
q q q q
magnitude of (1.25 ) k 2
. Charge C is pushed to the right by B with a force of k 2
d d

, Chapter 18 Answers to Focus on Concepts Questions 1173


q q
and pulled to the left by A with a force of k 2
. Since these two forces have different
( 2d )
q q
directions, the net force acting on C has a magnitude of ( 0.75 ) k .
d2

6. (b) According to Coulomb’s law, the magnitude of the force that any one of the point
q q
charges exerts on another point charge is given by F = k 2 , where d is the length of
d
each side of the triangle. The charge at B experiences a repulsive force from the charge
at A and an attractive force from the charge at C. Both forces have vertical components,
but one points in the +y direction and the other in the −y direction. These vertical
components have equal magnitudes and cancel, leaving a resultant that is parallel to the
x axis.

kq
7. (c) The electric field created by a point charge has a magnitude E = and is inversely
r2
proportional to the square of the distance r. If r doubles, the charge magnitude must
increase by a factor of 22 = 4 to keep the field the same.

8. (b) To the left of the positive charge the two contributions to the total field have opposite
directions. There is a spot in this region at which the field from the smaller, but closer,
positive charge exactly offsets the field from the greater, but more distant, negative
charge.

9. (a) The electric field has a greater magnitude where the field lines are closer together.
They are closest together at B and farthest apart at A. Therefore, the field has the
greatest magnitude at B and the smallest magnitude at A.

10. (d) In a conductor electric charges can readily move in response to an electric field. In
A, B, and C the electric charges experience an electric field and, hence, a force from
neighboring charges and will move outward, away from each other. They will rearrange
themselves so that the electric field within the metal is zero at equilibrium. This means
that they will reside on the outermost surface. Thus, only D could represent charges in
equilibrium.

11. 0.45 N·m2/C

12. (a) The ball is electrically neutral (net charge equals zero). However, it is made from a
conducting material, so it contains electrons that are free to move. The rod attracts some
of these (negative) electrons to the side of the ball nearest the rod, leaving the opposite
side of the ball positively charged. Since the negative side of the ball is closer to the
positive rod than the positive side, a net attractive force arises.

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