HW #1 Solutions
Q6) The net charge on a conductor refers to the difference between the number of
protons and the number of electrons. Neutral: difference =0; Positive: more protons than
electrons; Negative: more electrons than prot
HW #5 Solutions
Q5) The outlets in a double outlet are connected in parallel. Both outlets deliver the
same potential difference of 120 VAC and they operate independently. A lamp plugged
into one outlet can be turned on and
HW 4 Solutions
P32) Q = CV, so V = Q/C=16.510-8 C/950010-12 F = 17 V.
E = V/d. Know d, but need V. Use Q = CV so V = Q/C = 7210-6 C/0.8010-6 F = 90 V.
Thus E = 90 V/0.002 m = 45,000 V /m.
PE = 1/2CV2 = 0.5 220010-12 F (65
HW #3 Solutions
Ch 17 Questions
Q6) Yes. A particle with a negative charge moving from lower potential
to higher potential will have its potential energy decrease. For example
suppose the a charge of -3 C moves from a potential of
HW #2 Solutions
Q14) A negative charge can be used to determine the electric field. The
magnitude of E is still
, but the direction of E is opposite to the direction
of the force on the negative charge.
Q18) The e
HW #6 Solutions
Q19) No, the total energy supplied by the battery is not stored in the capacitor. Some of the
energy is dissipated as heat, as the charge flows from the battery to the capacitor through the
resistor. Recall f
HW #7 Solutions
The kinetic energy will stay the same. The force on the charged particle due to the
magnetic field is perpendicular to the direction the particle is traveling and therefore no work is
done (W = Fdcos; = 9
HW 12 Solutions
1) Ch23 P31
The angle between the light beam and the surface of the pool is given by tan = 1.3/2.7.
This gives = 25.7. The angle of incidence is 90 - 25.7 = 64.3. Using Snells law, the
angle of refraction is gi
HW 11 Solutions
Ch. 23 P72
a) Image 1: The closest image is the one you see looking in any single mirror. Since
you are 1.5 m from the mirror, the image appears 1.5 m behind the mirror. Thus it is 3.0
m from you.
Image 2: Th
HW 10 Solutions
1) a) C =
. A = r2 = 0.0038 m2. D = 0.025 m. C = 1.36 10-12 F.
b) Q = CV = 3.4 10-10 coulombs.
!c) E = V/d = 250/0.025 = 10,000 V/m.
d) Flux through left side = EA =10,000 V/m (0.01m)2 = 3.1 Nm2/C. E is into
HW #8 Solutions
The current moves through RB from right to left. Using the right hand rule for
solenoids, the left side of coil B is a north pole. Over by coil A, the field due to coil B is pointing
from right to left
HW #9 Solutions
1) a) FT = mg = .00444 kg 9.80 = 0.0435 N
b) Since the tension decreases, the force on the sphere due to the electric field must be up.
Since the sphere has a negative charge, the field must be down. Thus the top pl