223
Chapter 26
A capacitor is a device that one can use to store an electric potential energy.
(Spring: mechanical potential energy) Useful in electronics and
microelectronics
224
(C: capacitance SI: farad)
225
226
The plates of a parallel-plate capacitor are separated by a distance d=1.0mm. What must be the plate
area if the capacitance is to be 1.0F?
Sol:
A=
Cd
ε0
= 1.1X108 m2
(large!)
227
228
229
230
231
Energy density μ: The potential energy per unit volume
232
What is the potential energy of the two-capacitor system in problem 26-4 before and after switch S is
closed.
1
Ui= C1V02 = 70.4 μJ
2
1
1
Uf= C1V 2 + C 2V 2 = 20.0μJ
2
2
(smaller!)
Heat
(c) What is the radius R0 of an imaginary spherical surface such that half of the stored potential energy
lies within it?
(a)
233
(b)
(c)
∫
R0
R
dU =
1 ∞
dU
2 ∫R
dU = (u )(4πr 2 )(dr )
q 2 dr
dU =
8πε 0 r 2
∫
R0
R
dr 1 ∞ dr
=
r 2 2 ∫R r 2
1 1
1
−
=
R R0 2 R
R=R0=13.7cm
κ : a numerical factor
234
Water,etc.
235
236
237
(a)
(b)
(c)
(d)
(e)
238
(f)
Exercises:17,29,41
239
Chapter 27
Electric currents: Charges in motion
240
Current is a scalar.
For historical reasons, a current arrow is drawn in the direction in which positive
charge carriers would move, even if the actual charge carriers are negative and
move in the opposite direction.
For a constant current
(SI unit: A/m2)
Electrons tend to drift with a drift speed νd in the direction opposite to the
current.
241
(a)
Uniform current densityÆ
242
(b)
One end of an aluminum wire whose diameter is 2.5mm is welded to one end of a copper wire whose
diameter is1.8mm. The composite wire carries a steady current I of 17ma.
(a) What is the current density in each wire?
(b) What is the drift speed of the conduction electrons in the copper wire? Assume that, on the average,
each copper atom contributes one conduction electron.
(a)
Jcu=
17 mA
d
π ( )2
2
=6.7X103 A/m2
J A1 AA1
=
⇒ 3.5 X 10 3 A / m 2 = J A1
J cu
Acu
(b)
243
νd =
6.7 X 10 3 A / m 2
= 4.9 X 10 −7 m / s = 1.8mm / h
N Aρ
(
)(1.60 X 10 −19 )
M
Consider a strip of silicon has a rectangular cross section with width w=3.2mm
and height h=250μm, and through which there is a uniform current I of 5.2ma. The
silicon is an n-type semiconductor, having been “doped” with a controlled
phosphorus impurity. N=1.5X1023m-3
(a) What is the current density in the strip?
Sol:
J=i/wh=6500A/m2
(b) What is the drift speed?
Sol:
vd=J/ne=27cm/s
R: resistance
244
ρ: resistivity
245
246
( pn junction)
or
If an electron of mass m is placed in an electric field of magnitude E,
On the average
( : the mean free time)
247
(a)
(b)
248
A wire of length L=2.35m and diameter d=1.63mm carries a current I of 1.24A. The wire dissipates
electrical energy at the rate P of 48.5mW. of what is the wire made?
Sol:
4i 2 ρL
P= 2
πd
Æρ=2.80x10-8Ω．M
Æ
Al
The valence band: the highest band that is occupied by electrons.
Superconductors: Hg ,.....
High temperature superconductivity : Paul chu,...
Exercises:25,31,41
249
Chapter 28
A “charge pump”: a device that by doing work on the charge carriers maintains a
potential difference between a pair of terminals.
Æan emf device (electromotive force)
Def ξ
=
dW
dq
Ideal emf device: no internal resistance
Real emf device: (Battery) has internal resistance to the internal movement of
charge
250
251
252
(Kirchhoff’s current law)
253
254
(a)
255
(b)
(c)
256
Æ
Q=CV
Def. τ= RC : (the time constant )
t=τ
Æ
257
Æ
K=?
Initial condition
A capacitor of capacitance C is discharging through a resistor of resistance R
(a)In terms of the time constant τ=RC, when will the charge on the capacitor be half its
initial value?
258
(b)When will the energy stored in the capacitor half its initial value?
(c)At what rate PR is thermal energy produced in the resistor during the discharging process?
At what rate Pc is stored energy lost by the capacitor during the charging process
(a)
q = q0 e
−
t
RC
t
−
1
q 0 = q0 e RC
2
t
−
t
1
ln = ln(e RC ) = −
RC
2
T=-ln(1/2)(RC)=0.69τ
(b)
2
2t
2t
−
q 2 q 0 − RC
=
e
= U 0 e RC
U=
2C 2C
2t
−
1
U 0 = U 0 e RC
2
ln(1/2)=-2t/RC
t=0.35τ
(c)
t
q −
PR=i R= [ 0 e RC ] 2 R
RC
2
2
2t
−
q
= 0 2 e RC
RC
2t
2t
2
2t
−
−
2U −
2q
d
Pc=dU/dt= (U 0 e RC ) = − 0 e RC = − 0 2 e RC
dt
RC
RC
PR+PC=0
Exercises:31,43
259
Chapter 29
A charged plastic rod produces a vector field – the electric field E in the space around it.
A magnet produces a vector field – the magnetic field B in the space around it.
Setting up magnetic fields
(1) Moving electrically charged particles, such as a current in a wire
(2) Elementary particles such as electrons have an intrinsic magnetic field around
them.
Def.
--Using the right hand rule.
SI unit: Tesla (T)
260
261
(small)
a=?
(large)
2
(K<<mc ) relativistic
262
We have
263
Hall potential difference
264
265
266
267
Copper: screens the electric field.
(a) relativity
(b) large r
In vector form:
268
Figure 29-21 shows a length of wire with a central semicircular arc, placed in a uniform
magnetic field b that points out of the plane of the figure. If the wire carries a current i ,
what resultant force F acts on it?
269
dL
2
dFsinθ
R
θ
1
0
3
dF 0
Sol:
F1=F3=iLB
dF=iBdL=iB(Rdθ)
π
π
0
0
F2 = ∫ dF sin θ = ∫
(iBRdθ ) sin θ = iBR ∫
F= F1+ F2+F3 =2iB(L+R)
Electric motor
π
0
sin θdθ = 2iBR
270
Right-hand rule
271
Def
(magnetic dipole)
ΔU=(+μB)-(- μB)=2μB
Exercises:19,35,41,55