C &S Chapter 6 Notes
Where are the charges that create E inside
the wires that push the electrons thru the wire? On the outside of the wire. See figure page 210 & 211.
RBD #1.(6.2.1) It is variation of charge build
up along the outside of the wires that produces the E field that drives the
electrons thru the wires.
Why doesn't the current in the wire stop
or change when the wire is bent?
See fig pg 216.
RBD #2. 6.5
Resistors in the circuit (Pg 217). Recall i = nAv = nAuE.
Since i must be the same everywhere in a
series circuit, and n and u are the same everywhere in the wire, speed v must be greater in narrower areas.
iw = naw vw
= ir = nar vr
naw uEw = nar
uEr since Aw > Ar, Er > Ew
RBD#3. Battery maintains charge separation on the plates, and
hence the net electric field in the circuit. Enet = Er +
Ew = constant.
RBD#4. 6.6 Work done by the battery = FNC x distance between the
battery plates = FNC Ÿ s = Work done by the Electric field in the wires.
FNC Ÿs = FCLw + FCLr
= eEw Lw + eEr Lr
Page 221: About batteries:
About
6.6.1a E = Q/AŽo See
Page 143
RBD#5 6.7. Applications
of theory. Page 224. Note that Ewire
<< Eres !!
RBD#6. Page 226 Bulbs in parallel : l1 = l2
length of the wire in the bulbs, so both bulbs feel the same E, so FNC
Ÿs = El1 = E l2.
Since i = nAuE, i1 = i2 for identical bulbs.
RBD#7.
Page 227: 2 bulbs in series : ilong
= n1Along u Elong = iround = n2Around
uElong , which leads to Eround = (A1/A2)
Elong , hence , if A1 > A2, then Eround > Elong.
RBD#8. Pg 228: Long and round bulbs in series:
Why doesn't the round bulb glow?
Er = (AL/Ar) EL
<< EL. The work
done in pushing electrons thru bulbs determines how much they glow. Wr = e Er
L <<e ELL, (L = length of the filiments.)
RBD #9: Pg 229: more than 2 batteries, use wire, not bulbs, you'll burn out
the bulbs.
Also,
for two batteries, E = Qs/(2Re0),
with s twice as large for two batteries.