HW #3 Solutions
16.19 Using conservation of energy, we have KEi +PEi = KEf + PEf.
qαqgold
ButPEi = k r
, and ri = . Thus, PEi = 0.
i
Also KEf = 0 (vf = 0 at turning point), so PEf = KEi, or
qαqgold
1
k r
= 2 mαvα2, and
min
2kqαqgold 2(8.99 x 109 N m2/C2)(2e)(79e)
=
= 2.8 x 10-14 m.
rmin =
(6.6 x 10-27 kg)(2 x 107 m/s)2
mαvα2
16.23 (a)
If d is doubled while A and Q remain constant, then when d doubles, C is
reduced by a factor of 2. Thus, if Q remains constant, then
Q2
Q1
Q1
∆V2 = C = 1
= 2 C , or ∆V2 = 2 ∆V1 = 800 V.
2
1
C
2 1
(b) If d is doubled the capacitance reduced by a factor of 2. Thus, if ∆V is held
constant:
1
1
1 
Q2 = C2∆V2 = 2 C1 (∆V1) = 2 C1 ∆V1= 2 Q1, or the charge must be


halved.
16.29 (a)
Using the rules for combining capacitors in series and in parallel, the circuit
is reduced in steps as shown below. The equivalent capacitor is shown to be
a 2 µF capacitor.
4 µF
3 µF
a
b
2 µF
12 V
Figure 1
6 µF
c
→
a
2 µF
3 µF
b
c
12 V
Figure 2
→
a
c
12 V
Figure 3
(b) From Fig. 3:
Qac = Cac∆Vac = (2.0 µF)(12 V) = 24 µC.
From Fig. 2:
Qab = Qbc = Qac = 24 µC
Thus, the charge on the 3.0 µF capacitor is: Q3 = 24 µC.
We now find the potential differences between the points indicated in Fig. 2:
Qab 24 µC
Qbc 24 µC
∆Vab = C =
= 4.0 V, and ∆Vbc = C =
= 8.0 V
ab 6.0 µF
bc 3.0 µF
Finally, usingFig. 1:
Q4 = C4∆Vab = (4.0 µF)(4.0 V) = 16 µC,
and
Q2 = C2∆Vab = (2.0 µF)(4.0 V) = 8.0 µC.
16.36 The initial charge on the 10.0 µF capacitor is Q = (12.0 V)(10.0 µF) = 120 µC.
When the two capacitors are connected in parallel, the voltage across each
capacitor is 3.0 V. Thus, the charge remaining on the 10.0 µF capacitor is:
Q10 = C10∆V' = (10.0 µF)(3.0 V) = 30.0 µF.
The rest of the 120 µC initial charge is now stored on the new capacitor (i.e., QC
= 120 µC - 30 µC = 90 µC). Thus, the second capacitor has a capacitance of:
QC 90 µF
C=
=
= 30 µF.
∆V' 3.0 V
16.39 (a)
Each capacitor has voltage V, so:
1
1
W = 2 (C1 + C2) ∆V2 = (25 x 10-6 F + 5.0 x 10-6 F)(100 V) 2 = 0.15 J.
2
(b) When in series the equivalent capacitance is 4.167 µF, and
1
0.15 J = 2 (4.167 x 10-6 F) ∆V'2, or ∆V' = 270 V.
17.14
∆V
12 V
(a) The resistance of the wire is: R = I = 0.40 A = 30 Ω, and
RA 30 Ω π(4.0 x 10-3 m)2
(b) ρ =
= 4.7 x 10-4 Ωm
L =
3.2 m
17.33
(a) The energy W = power times the time used. Thus,
W = Pt = (90 J/s)(3600 s) = 3.2 x 105 J.
(b) The power consumed by the color set is,
P = ∆VI = (120 V)(2.5 A) = 300 W.
W 3.24 x 105 J
= 1.1 x103 s = 18 min.
Thus,
t = P = 300 W
17.47
The resistance of the 4.0 cm length of wire between the feet is:
ρL ( 1.7 x 10-8 Ω m) ( 0.040 m)
= 1.8 x 10-6 Ω.
R= A =
π(1.1 x 10-2 m)2
The voltage between the feet is:
∆V = IR = (50 A)(1.8 x 10-6 Ω) = 9.0 x 10-5 V = 90 µV
17.49
(a) From the definition of current, we see that:
∆Q = I(∆T).
From this, we see that the charge can be found by finding the total area
under a curve of I versus t. Thus,
Q = (area of rectangle A1) + 2(area of triangle A2)
+ (rectangular area A3).
(1 s)(4 A)
Q = (2 A)(5 s) + 2
 + (1 s)(4 A) = 10 C + 4 C + 4 C = 18 C.
2


∆Q
18 C
(b) The constant current would be: I =
= 5 s = 3.6 A.
∆t