Brock University
Physics 1P22/1P92
Winter 2014
Dr. D’Agostino
Solutions for Tutorial 6: Electric Potential (Chapter 21) and Current and Resistance (Chapter 22)
1. A parallel-plate capacitor with plate separation d is connected to a battery that has
potential difference ∆V . Without breaking any of the electrical connections, insulating
handles are used to increase the plate separation to 2d while keeping the plates parallel.
(a) Does the potential difference between the plates change as the separation between
the plates increase? If so, by what factor? If not, explain why not. [2 points]
(b) Does the capacitance change as the plate separation increases? If so, by what
factor? If not, explain why not. [2 points]
(c) Does the charge on the capacitor change as the plate separation increases? If so,
by what factor? If not, explain why not. [2 points]
Solution: (a) Because the battery remains connected to the capacitor as the plates
are moved, the potential difference across the plates remains constant.
(b) According to the formula
κε0 A
d
when the distance between the plates increases, the capacitance of a parallel-plate
capacitor increases. In fact,
C=
C2
C1
C2
C1
C2
C1
C2
C1
C2
C1
C2
C1
=
=
=
=
=
=
κε0 A/d2
κε0 A/d1
1/d2
1/d1
1 d1
·
d2 1
d1
d2
d1
2d1
1
2
(c) Based on the relation Q = C∆V and the arguments in Parts (a) and (b), the charge
on each plate changes:
Q2
C2 ∆V
=
Q1
C1 ∆V
Q2
C2
=
Q1
C1
Q2
1
=
Q1
2
Because the capacitance decreases while the potential difference across the plates remains constant, the charge on each plate decreases by the same factor as the capacitance.
2. You need to construct a 50 pF capacitor. You plan to cut two L × L metal squares
and place a dielectric material between them. The thinnest dielectric you have is 0.1
mm thick, and its dielectric constant is 1.2. Determine L. [4 points]
Solution:
κε0 A
d
Cd
A=
κε0
Cd
L2 =
κε0
(50 × 10−12 F)(0.1 × 10−3 m)
L2 =
(1.2)(8.85 × 10−12 C2 /N·m2 )
L2 = 4.71 × 10−4 m2
L = 2.17 × 10−2 m
L = 2.2 cm
C=
3. The wires in the figure are all made of the same material. Rank in order, from largest
to smallest, the resistance of each wire. Explain. [5 points]
Solution: Begin with the formula
ρL
A
Now replace the area A by the equivalent expression πr2 to obtain
ρL
R= 2
πr
Thus,
ρL
R1 = 2
πr
ρL
1 ρL
1
R2 =
=
·
=
· R1
π(2r)2
4 πr2
4
ρ2L
1 ρL
1
R3 =
= · 2 = · R1
2
π(2r)
2 πr
2
ρ2L
ρL
R4 =
= 2 2 = 2R1
2
πr
πr
ρL
ρ4L
R5 =
= 2 = R1
2
π(2r)
πr
R=
Thus,
R4 > R1 = R5 > R3 > R2
4. A 70 W electric blanket runs at 18 V.
(a) Determine the resistance of the wire in the blanket. [1 points]
(b) Determine the current in the wire. [1 points]
(c) If the wire is made of copper and is 20 m long, determine the radius of the wire.
[3 points]
Solution: (a) Because P =
V2
,
R
it follows that
V2
P
182
R=
70
R = 4.63 Ω
R=
(b) Because P = IV , it follows that
P
V
70
I=
18
I = 3.89 A
I=
(c)
ρL
A
ρL
R= 2
πr
ρL
2
r =
πR
1.7 × 10−8 Ω· m(20 m)
r2 =
π(4.63 Ω)
2
r = 2.338 × 10−8 m2
r = 1.53 × 10−4 m
r = 0.15 mm
R=