Transcript Chap. 17 Conceptual Modules Giancoli

ConcepTest 2.8 Capacitors
Capacitor C1 is connected across
a) C1
a battery of 5 V. An identical
b) C2
capacitor C2 is connected across
a battery of 10 V. Which one has
c) both have the same charge
d) it depends on other factors
the most charge?
+Q –Q
ConcepTest 2.8 Capacitors
Capacitor C1 is connected across
a) C1
a battery of 5 V. An identical
b) C2
capacitor C2 is connected across
a battery of 10 V. Which one has
the most charge?
c) both have the same charge
d) it depends on other factors
+Q –Q
Since Q = C V and the two capacitors are
identical, the one that is connected to the
greater voltage has the most charge,
which is C2 in this case.
ConcepTest 2.9a Varying Capacitance I
What must be done to
a) increase the area of the plates
a capacitor in order to
b) decrease separation between the plates
increase the amount of
c) decrease the area of the plates
charge it can hold (for
a constant voltage)?
d) either (a) or (b)
e) either (b) or (c)
+Q –Q
ConcepTest 2.9a Varying Capacitance I
What must be done to
a) increase the area of the plates
a capacitor in order to
b) decrease separation between the plates
increase the amount of
c) decrease the area of the plates
charge it can hold (for
a constant voltage)?
d) either (a) or (b)
e) either (b) or (c)
+Q –Q
Since Q = C V, in order to increase the charge
that a capacitor can hold at constant voltage,
one has to increase its capacitance. Since the
capacitance is given by C   0 A , that can be
d
done by either increasing A or decreasing d.
ConcepTest 2.9b Varying Capacitance II
A parallel-plate capacitor
a) the voltage decreases
initially has a voltage of 400 V
b) the voltage increases
and stays connected to the
c) the charge decreases
battery. If the plate spacing is
now doubled, what happens?
d) the charge increases
e) both voltage and charge change
+Q –Q
ConcepTest 2.9b Varying Capacitance II
A parallel-plate capacitor
a) the voltage decreases
initially has a voltage of 400 V
b) the voltage increases
and stays connected to the
c) the charge decreases
battery. If the plate spacing is
now doubled, what happens?
d) the charge increases
e) both voltage and charge change
Since the battery stays connected, the
voltage must remain constant ! Since
C   0 A when the spacing d is doubled,
d
the capacitance C is halved. And since
Q = C V, that means the charge must
decrease.
Follow-up: How do you increase the charge?
+Q –Q
ConcepTest 2.9c Varying Capacitance III
A parallel-plate capacitor initially has
a potential difference of 400 V and is
then disconnected from the charging
battery. If the plate spacing is now
doubled (without changing Q), what
is the new value of the voltage?
+Q –Q
a) 100 V
b) 200 V
c) 400 V
d) 800 V
e) 1600 V
ConcepTest 2.9c Varying Capacitance III
A parallel-plate capacitor initially has
a potential difference of 400 V and is
then disconnected from the charging
battery. If the plate spacing is now
doubled (without changing Q), what
is the new value of the voltage?
Once the battery is disconnected, Q has to
remain constant, since no charge can flow
either to or from the battery.
Since
C   0 A when the spacing d is doubled, the
d
capacitance C is halved. And since Q = C V,
that means the voltage must double.
a) 100 V
b) 200 V
c) 400 V
d) 800 V
e) 1600 V
+Q –Q
ConcepTest 2.11a
Capacitors I
a) Ceq = 3/2 C
What is the equivalent capacitance,
b) Ceq = 2/3 C
Ceq , of the combination below?
c) Ceq = 3 C
d) Ceq = 1/3 C
e) Ceq = 1/2 C
o
Ceq
o
C
C
C
ConcepTest 2.11a
Capacitors I
a) Ceq = 3/2 C
What is the equivalent capacitance,
b) Ceq = 2/3 C
Ceq , of the combination below?
c) Ceq = 3 C
d) Ceq = 1/3 C
e) Ceq = 1/2 C
The 2 equal capacitors in series add
o
up as inverses, giving 1/2 C. These
are parallel to the first one, which
Ceq
add up directly. Thus, the total
equivalent capacitance is 3/2 C.
o
C
C
C
ConcepTest 2.11b
Capacitors II
How does the voltage V1 across
a) V1 = V2
the first capacitor (C1)compare
b) V1 > V2
to the voltage V2 across the
c) V1 < V2
second capacitor (C2)?
d) all voltages are zero
C2 = 1.0 mF
10 V
C1 = 1.0 mF
C3 = 1.0 mF
ConcepTest 2.11b
Capacitors II
How does the voltage V1 across
a) V1 = V2
the first capacitor (C1)compare
b) V1 > V2
to the voltage V2 across the
c) V1 < V2
second capacitor (C2)?
d) all voltages are zero
The voltage across C1 is 10 V.
The combined capacitors
C2+C3 are parallel to C1. The
voltage across C2+C3 is also
10 V. Since C2 and C3 are in
series, their voltages add.
Thus the voltage across C2
and C3 each has to be 5 V,
which is less than V1.
C2 = 1.0 mF
10 V
C1 = 1.0 mF
C3 = 1.0 mF
Follow-up: What is the current in this
circuit?
ConcepTest 2.11c
How does the charge Q1 on the first
Capacitors III
a) Q1 = Q2
b) Q1 > Q2
capacitor (C1) compare to the
charge Q2 on the second capacitor
c) Q1 < Q2
(C2)?
d) all charges are zero
C2 = 1.0 mF
10 V
C1 = 1.0 mF
C3 = 1.0 mF
ConcepTest 2.11c
How does the charge Q1 on the first
Capacitors III
a) Q1 = Q2
b) Q1 > Q2
capacitor (C1) compare to the
charge Q2 on the second capacitor
c) Q1 < Q2
(C2)?
d) all charges are zero
We already know that the
C2 = 1.0 mF
voltage across C1 is 10 V
and the voltage across C2
and C3 each is 5 V. Since Q
= CV and C is the same for
all the capacitors, then since
V1 > V2 therefore Q1 > Q2.
10 V
C1 = 1.0 mF
C3 = 1.0 mF