Download Voltage, Electric Potential, and Capacitance Exercises and more Assignments Chemistry in PDF only on Docsity!
Chapter 19 Problems
19.1 Electric Potential Energy: Potential Difference
1. Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential
difference? voltage = potential of point with respect to a reference potential point
2. If the voltage between two points is zero, can a test charge be moved between them with zero net work
being done? Can this necessarily be done without exerting a force? Explain.
A test charge can be moved bw them with zero net work being done and Yes, the net force is zero so if it is
already moving it will keep moving.
3. What is the relationship between voltage and energy? More precisely, what is the relationship between
potential difference and electric potential energy?
Voltage = potential energy per unit charge (PE/q) à change in Voltage = W/q (W = change in PE)
Potential difference is electric potential energy per unit charge.
4. Voltages are always measured between two points. Why? Voltage is a difference in electric potential
5. How are units of volts and electron volts related? How do they differ? An electron volt is a unit of energy: it
is the energy gained by an elementary charge when it is accelerated across a potential difference of one volt.
Problems:
1. Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron)
accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen
ion to be 1.67×
2. An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and
produce x rays. Non-relativistically, what would be the maximum speed of these electrons?
o
kE=Imu
Emde
'
= (^) # multi
÷=fm#=×
- 11 ×10-
e
✓
'
may
=
I - 185 × 108
mfs
KE =
ImV2max
qV=Imv
=
. Ge
-_ 1.6×10-
m
speed
, V
= 4 × 104
Me
=
9.11×10-
PE
19.2 Electric Potential in a Uniform Electric Field
6. Discuss how potential difference and electric field strength are related. Give an example. Electric field
strength is change in electric potential per unit length. It is a vector that is perpendicular to the equipotential.
7. What is the strength of the electric field in a region where the electric potential (Voltage) is constant?
E = dV/ dx. Electric field strength is given by the gradient of electric potential à if voltage is constant, electric
field strength will be zero
8. Will a negative charge, initially at rest, move toward higher or lower potential? Explain why.
Electric current flows from higher to lower potential. As the direction of moving electrons is opposite to that
of the current, the electrons flow from the region of lower potential to higher potential.
W = - q(change in V) à final potential > initial potential
Negative charge will move towards the higher potential.
Problems:
14. What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and
having a potential difference (voltage) between them of 1.50×104V?
15. The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50×104V/m.
(a) What is the potential difference between the plates?
(b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that
plate (and 3.00 cm from the other)?
16. How far apart are two conducting plates that have an electric field strength of 4.50×103V/m between
them, if their potential difference is 15.0 kV?
O
f-
IHO
m
E-
VI
= I^ -5× 106
Ulm
✓ =^ 1.5× 104 U
O
E -
7.5× 104 Ulm
V
Ed =
3000 V
d = 4 ×^10
m
E
D= 1 ×10-
V= 7-^
Sov
E-^ 4.5× 103 ✓
✓
= 15kV
= 15000W
E
f- YE
= (^) 3.33M
26. (a) A sphere has a surface uniformly charged with 1.00 C. At what distance from its center is the potential
5.00 MV?
(b) What does your answer imply about the practical aspect of isolating such a large charge?
27. How far from a 1.00 μC point charge will the potential be 100 V?
At what distance will it be 2.00×102V?
28. What are the sign and magnitude of a point charge that produces a potential of – 2.00 V at a distance of
1.00 mm?
29. If the potential due to a point charge is 5.00×102V at a distance of 15.0 m, what are the sign and
magnitude of the charge?
30. There is a 2.5 nC charge on the x-axis at x = 0.150 m and a - 5.0 nC charge on the y-axis at y = 0.10 m. find
the electric potential at the origin.
f- KRI r
O
G-
1C
✓
=
KCIE )
✓
= -5× 106
✓ ¥^
= 1800 m
IC
charge acquired by sphere arge
the distance is large
potential
-_ (^) 5mV → Impractical
O
✓
= (^) IOOV
Q
lxio-6C
V' too -_kCt×f
r=9Om
2 ×102=
I
r=
45ns
V
V=
=
KG of
-_
BC
(1×10-3)
5 ×102=
q=
8.33×10-
C
13M
=
=
fo
v ,
= kl-5Xt
=
of
-_ (^) 2. 5h C^ =^
2.5× 10 -9C 0.
T.rs
? m
✓
=
-1cg
Vnet
( (^) 2- Jrc )
Vz=kl25X
= , go
r
not vector
V=K&÷
KEI
X Nz
0.10M l
Snc )
(
-5×10-9 )
=k
k
"IY:
I S
O
(2-5h
C )
W=
GZOV
Gi
Gz
( VB
VA )
A
B
=qz(k8%
)
=
KGW
z
to .o^
=
3.15× 10 -7J
B
Y
U
=k9r
-¥.,z
VB =^ k
(-2.2410-9)-
x
VA
= k
(-2-2×-10-9)
O
Vrs
A.
2 ×10-9)
(
t.su
V
B)
How
much
W
done
by
e
Id in^
moving
n
from (^) A to B
A
-1 B
INI
potential
l
W
of
OV = (^) -10-9 ( 128 )
37. Sketch the equipotential lines for the two equal positive charges shown in Figure 19.26. Indicate the
direction of increasing potential.
38. Figure 19.27 shows the electric field lines near two charges q1 and q2, the first having a magnitude four
times that of the second. Sketch the equipotential lines for these two charges, and indicate the direction of
increasing potential.
39. Sketch the equipotential lines a long distance from the charges shown in
Figure 19.27. Indicate the direction of increasing potential.
q1 = negative (electric field ending on it); q2= positive (electric field lines originate
from)
At long distance, ONLY effective charge, - 4q + q = - 3q is visible à electric field
lines will be like that of - 3q charge particle at a long distance
- potential is greater as one approaches the POS charge and become leaset (MOST
NEG) near negative charge è direction of increasing potential will be away from
neg charge
19.5 Capacitors and Dielectrics
14. Does the capacitance of a device depend on the applied voltage? No What about the charge stored in it?
Yes, the more voltage, the more charge. The capacitance gives the ratio.
15. Use the characteristics of the Coulomb force to explain why capacitance should be proportional to the
plate area of a capacitor. Increasing the plate area allows the like charges on each plate to spread out more
without moving them farther from the opposite charges on the other plate
Similarly, explain why capacitance should be inversely proportional to the separation between plates. As the
plates move apart, the opposite charges on the plates feel less attractive force, reducing the charge that builds
up for a given voltage.
16. Give the reason why a dielectric material increases capacitance compared with what it would be with air
between the plates of a capacitor. The dielectric molecules polarize, reducing the electric field for a given
charge. What is the independent reason that a dielectric material also allows a greater voltage to be applied to
a capacitor? It reduces the chance that the capacitor will short circuit by sparking (The dielectric thus
increases C and permits a greater V.)
E- (^) T#
←to
F
→ PE
W PE
AKE
19.6 Capacitors in Series and Parallel
21. If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series
or parallel? Parallel Explain. To increase stored energy for a given voltage, you must increase the capacitance.
Connecting them in parallel is similar to increasing the plate area.
Ecap = ½(Cnet x V^2)
voltage applied across a network is same = energy stored by capacitor network is directly proportional to
Cnet
Problems:
57. Find the total capacitance of the combination of capacitors in Figure 19..
58. Suppose you want a capacitor bank with a total capacitance of 0.750 F and you possess numerous 1.50 mF
capacitors. What is the smallest number you could hook together to achieve your goal, and how would you
connect them?
59. What total capacitances can you make by connecting a 5.00 μF and an 8.00 μF capacitor together?
60. Find the total capacitance of the combination of capacitors shown in Figure 19..
C , z^
=
12.5μF
3--42.5+40.3=
293 μF
C total
= o.^ -7 (^) SF
Ceq=
n C
C -^ c.^ 5mF xco^
n'
parallel (^) Ceq
= (^) Jtf =
13 μF
series 's
Ys
t Yq=
Ceqa
MF
Chet
=
Cut
R (^) -
=
2.79μF
series
%
,.
( ie
-^ 0.
E capacitance
= I
(Cne^ x
x U2^ )
61. Find the total capacitance of the combination of capacitors shown in Figure 19..
19.7 Energy Stored in Capacitors
22. How does the energy contained in a charged capacitor change when a dielectric is inserted, assuming the
capacitor is isolated and its charge is constant? Does this imply that work was done? The capacitance
increases so, if the charge does not change, the voltage decreases, decreasing the energy. Work is done in
inserting the dielectric and this goes into polarizing the molecules.
23. What happens to the energy stored in a capacitor connected to a battery when a dielectric is inserted?
Was work done in the process? If dielectric is inserted bw plates of capacitor, keeping the voltage constant,
the energy stored in capacitor increases. The charge increases so the energy increases.
The work done in inserting dielectric raises the PE of capacitor.
Problems:
63. (a) What is the energy stored in the 10.0 μF capacitor of a heart defibrillator charged to 9.00×103V?
(b) Find the amount of stored charge.
64. In open heart surgery, a much smaller amount of energy will defibrillate the heart. (a) What voltage is
applied to the 8.00 μF capacitor of a heart defibrillator that stores 40.0 J of energy?
(b) Find the amount of stored charge.
( AB
=
E
C
CABC
=
D
C. ABCD
= (^) 1.37+8=9.
F
A B CEF
= (^) 2.
Ceq= 11.43μF
PE
ICUZ =
I Cioxco
6) ( 9 × 1035 = 405 J
G
( go^ yo^
9 × 103 )
= 0.09C
405=12 ( 8 ×10-6 )^ ✓
Z
✓
=
€μo
=
V
G CV^
=
( 8 ×10-^
c) (^ 3162.28W
=
0.025C
