25.17. Model: Charges A, B, and C are point charges.
Visualize: Please refer to Figure EX25.17.
Solve: The force on B from charge A is directed downward since two negative charges repel.
Coulombs law gives the magnitude of the force as
FA on B = K
qA qB
r
bee29400_ch11_600-689.indd Page 620
11/25/08
5:46:40 PM user-s173
/Volumes/204/MHDQ077/work%0/in
SAMPLE PROBLEM 11.4
A ball is thrown vertically upward from the 12-m level in an elevator shaft
with an initial velocity of 18 m/s. At the same instant an ope
30.3. Solve: The number of electrons crossing a cross-sectional area of a wire is the electron current i.
2
i = nAvd = (6.0 10
28
$ 1.6 103 m '
4
19 1
m ) &
) (2.0 10 m/s) = 2.4 10 s
2
%
(
3
The electron density n for aluminum is taken from Table 30.1. Th
28.1. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform electric
field.
Visualize:
The figure shows the before-and-after pictorial representation. The proton has an initial speed vi = 0 m/s and a
final speed
Physics 133 Schedule - Winter 2014 (White)
Date
Reading (before class)
Jan 6
25.1-3
Jan 8
25.4
Jan 10
quiz
25.5
2 Ch 26: Electric Field
Jan 13
26.1-2
Jan 15
26.3-5
Jan 17
quiz
26.6-7
3 Ch 27: Gauss Law
Jan 21 (Tues = Mon)
27.1-3
Jan 22 TIP#1
27.4-6
Jan 24
32.4. Model: The magnetic field is that of a moving charged particle.
Visualize:
The first point is on the x-axis, with a = 90. The second point is on the z-axis, with b = 0, and the third point
is in the yz plane with c = 45.
Solve: (a) Using the Biot-Sa
31.4. Model: Assume ideal connecting wires and an ideal battery for which Vbat = .
Visualize: Please refer to Figure EX31.4. We will choose a clockwise direction for I. Note that
the choice of the currents direction is arbitrary because, with two batterie
Week 2 HW Solutions (Chapter 26)
26.4. Model: The electric field at the point is found by superposition of the fields due to
the two charges located on the y-axis.
Visualize: The electric field due to the positive charge q1 at the point is away
33.1. Visualize:
To develop a motional emf the magnetic field needs to be perpendicular to both the velocity and the current, so
lets say its direction is into the page.
Solve: This is a straightforward use of Equation 33.3. We have
1.0 V
=
= 2.0 104 m/s
27.5. Model: The electric flux flows out of a closed surface around a region of space containing a net
positive charge and into a closed surface surrounding a net negative charge.
Visualize: Please refer to Figure EX27.5. Let A be the area of each of the