Download +C - UCSD Physics

Week 5
1. zero
Where an electric field line
crosses an equipotential
surface, the angle between the
field line and the equipotential is
2. between zero and 90°
3. 90°
4. not enough information given
to decide
1. zero
Where an electric field line
crosses an equipotential
surface, the angle between the
field line and the equipotential is
2. between zero and 90°
3. 90°
4. not enough information given
to decide
Since there is no change in V along the
equipotential surface
ΔV = −E⋅dr = 0
But this means that the vectors E and dr
are perpendicular to each other.
1. 1 Volt
2. 2 Volt
What is the electric potential
created by these three batteries
connected as shown?
3. 3 Volt
4. 4 Volt
5. 5 Volt
6. 6 Volt
7. 7 Volt
8. 8 Volt
1. 1 Volt
2. 2 Volt
What is the electric potential
created by these three batteries
connected as shown?
3. 3 Volt
4. 4 Volt
5. 5 Volt
6. 6 Volt
7. 7 Volt
8. 8 Volt
Which potentialenergy graph
describes this
electric field?
1
2
3
4
5
Which potentialenergy graph
describes this
electric field?
1
2
3
4
5
1. is the same as the direction of the
electric field at that point
The direction of the
electric potential gradient
at a certain point .....
2. is opposite to the direction of the
electric field at that point
3. is perpendicular to the direction of the
electric field at that point
4. not enough information given to decide
1. is the same as the direction of the
electric field at that point
The direction of the
electric potential gradient
at a certain point .....
2. is opposite to the direction of the
electric field at that point
3. is perpendicular to the direction of the
electric field at that point
4. not enough information given to decide
! = −∇V
!
E
Understand first a simpler analog:
topographic maps have “contour lines” that are the
locus of points such that ground elevation is constant:
h(!r) = constant
When you follow a path on a topographic map that crosses
contour lines, you are either climbing or descending. A path
running parallel to contour lines is flat.
DENSE contour lines mean STEEP terrain and SPARSE ones mean
FLAT terrain.
The “gradient” is in the direction of steepest ascent and its size
is the rate of ascent.
Which set of
equipotential
surfaces
matches this
electric field?
1.
2.
4.
3.
5.
Which set of
equipotential
surfaces
matches this
electric field?
1.
2.
4.
3.
5.
Example: Determine the electric field of a dipole, far away from the dipole
Use the result from last week:
x
cos θ
V = kp 2 = kp 3
r
r
(COB)
Ans:
r2 − 3x2
1 − 3 cos2 θ
Ex = −kp
= −kp
5
r
r3
3xy
sin θ cos θ
Ey = kp 5 = 3kp
r
r3
Three charged, metal spheres of
different radii are connected by
a thin metal wire. The potential
and electric field at the surface
of each sphere are V and E.
Which of the following is true?
1. V1 = V2 = V3 and E1 = E2 = E3
2. V1 = V2 = V3 and E1 > E2 > E3
3. V1 > V2 > V3 and E1 = E2 = E3
4. V1 > V2 > V3 and E1 > E2 > E3
5. V3 > V2 > V1 and E1 = E2 = E3
Three charged, metal spheres of
different radii are connected by
a thin metal wire. The potential
and electric field at the surface
of each sphere are V and E.
Which of the following is true?
1. V1 = V2 = V3 and E1 = E2 = E3
2. V1 = V2 = V3 and E1 > E2 > E3
3. V1 > V2 > V3 and E1 = E2 = E3
4. V1 > V2 > V3 and E1 > E2 > E3
5. V3 > V2 > V1 and E1 = E2 = E3
Capacitors
(slide here really to remind me of demo 1st)
A system consisting of t wo conductors (“plates”)
Now, equal magnitude,
opposite sign charge, ±Q , produces
field bet ween them
(a)
Hence, potential difference V
The more charge Q , the larger V
This is a linear relation (V
Q)
1
V = Q
C
–
–
+
–
–
–
–
+
+ +
(b)
+
+
The two conductors a and b are insulated
from each other, forming a capacitor. You
increase the charge on a to +2Q and
increase the charge on b to –2Q, while
keeping the conductors in the same
positions.
What effect does this have on the
capacitance C?
1. C is multiplied by a factor of 4
2. C is multiplied by a factor of 2
3. C is unchanged
4. C is multiplied by a factor of 1/2
5. C is multiplied by a factor of 1/4
The two conductors a and b are insulated
from each other, forming a capacitor. You
increase the charge on a to +2Q and
increase the charge on b to –2Q, while
keeping the conductors in the same
positions.
What effect does this have on the
capacitance C?
1. C is multiplied by a factor of 4
2. C is multiplied by a factor of 2
3. C is unchanged
4. C is multiplied by a factor of 1/2
5. C is multiplied by a factor of 1/4
You reposition the two plates of a
capacitor so that the capacitance
doubles.
If the charges +Q and –Q on the two
plates are kept constant in this
process, what happens to the potential
difference Vab between the two plates?
1. Vab is multiplied by a factor of 4
2. Vab is multiplied by a factor of 2
3. Vab is unchanged
4. Vab is multiplied by a factor of 1/2
5. Vab is multiplied by a factor of 1/4
You reposition the two plates of a
capacitor so that the capacitance
doubles.
If the charges +Q and –Q on the two
plates are kept constant in this
process, what happens to the potential
difference Vab between the two plates?
1. Vab is multiplied by a factor of 4
2. Vab is multiplied by a factor of 2
3. Vab is unchanged
4. Vab is multiplied by a factor of 1/2
5. Vab is multiplied by a factor of 1/4
Example: our demo
d is approx 0.5mm, and A is approx 1.0 m2
(COB)
C ≈ 1.8 × 10−8 F = 18 nF
so how much
charge did we
put in?
for 1 kV:
Q = CV ≈ 1.8 × 10−5 C
(This ignores the dielectric constant of the mica)
Capacitor C1 is connected
across a battery of 5 V. An
identical capacitor C2 is
1) C1
connected across a battery of 10
V. Which one has the most
charge?
+Q –Q
2) C2
3) both have the same charge
4) it depends on other factors
Capacitor C1 is connected
across a battery of 5 V. An
identical capacitor C2 is
connected across a battery of 10
V. Which one has the most
charge?
1) C1
2) C2
3) both have the same charge
4) 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.
What must be done to
a capacitor in order to
increase the amount of
charge it can hold (for
a constant voltage)?
1) increase the area of the plates
2) decrease separation between the plates
3) decrease the area of the plates
4) either (1) or (2)
5) either (2) or (3)
+Q –Q
What must be done to
a capacitor in order to
increase the amount of
charge it can hold (for
a constant voltage)?
1) increase the area of the plates
2) decrease separation between the plates
3) decrease the area of the plates
4) either (1) or (2)
5) either (2) or (3)
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
, that can be
done by either increasing A or decreasing d.
+Q –Q
A parallel-plate capacitor
initially has a voltage of 400 V
and stays connected to the
battery. If the plate spacing is
now doubled, what happens?
1) the voltage decreases
2) the voltage increases
3) the charge decreases
4) the charge increases
5) both voltage and charge change
+Q –Q
A parallel-plate capacitor
initially has a voltage of 400 V
and stays connected to the
battery. If the plate spacing is
now doubled, what happens?
1) the voltage decreases
2) the voltage increases
3) the charge decreases
4) the charge increases
5) both voltage and charge change
Since the battery stays connected, the
voltage must remain constant ! Since
when the spacing d is
doubled, 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
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
1) 100 V
2) 200 V
3) 400 V
4) 800 V
5) 1600 V
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
when the spacing d is doubled, the
capacitance C is halved. And since Q = C V,
that means the voltage must double.
1) 100 V
2) 200 V
3) 400 V
4) 800 V
5) 1600 V
+Q –Q
The textbook calculates the capacitance of
concentric spheres. Do another example:
L
co-axial cylindrical conductors
Use previous chapter result (also from homework):
! "
λ
b
V (a) − V (b) =
ln
2π#0
a
(COB)
2π"0 L
C=
ln(b/a)
Again, notice:
- only depends on geometry,
- increases with size of plates,
- increases as plates get close ( a → b )
a
b
Connecting circuit components
in Parallel and in Series
Some circuit component
Two circuit elements connected in series:
the charge of one necessarily flows into the other:
Two circuit elements connected in parallel:
the potential differences across them are necessarily equal:
Equivalent capacitance
of capacitors in parallel
C1
C1
C2
=
C1
C2
C1
C2
(COB)
(a)
(b)
C = C1 + C2
C2
Equivalent capacitance
of capacitors in series
C1
=
C2
C2
C2
+Q
+
+
+
+
+
+
+
+
C1
C1
–Q – – – – – – – –
(COB)
(b)
+Q
+
+
+
+
+
+
+
+
1
1
1 C2
=
+
C
C1
C2
E
–Q – – – – – – – –
E
Two capacitors are connected together
as shown.
1. Ceq = 18 µF
2. Ceq = 9 µF
3. Ceq = 6 µF
What is the equivalent capacitance of
the two capacitors as a unit?
4. Ceq = 4 µF
5. Ceq = 2 µF
Two capacitors are connected together
as shown.
1. Ceq = 18 µF
2. Ceq = 9 µF
3. Ceq = 6 µF
What is the equivalent capacitance of
the two capacitors as a unit?
4. Ceq = 4 µF
5. Ceq = 2 µF
Two capacitors are connected
together as shown.
1. 48 µC
2. 36 µC
3. 24 µC
If the charge on the 12–µF capacitor
is 24 microcoulombs (24 µC), what is
the charge on the 6–µF capacitor?
4. 12 µC
5. 6 µC
Two capacitors are connected
together as shown.
1. 48 µC
2. 36 µC
3. 24 µC
If the charge on the 12–µF capacitor
is 24 microcoulombs (24 µC), what is
the charge on the 6–µF capacitor?
4. 12 µC
5. 6 µC
A
Example: Find the equivalent
capacity of the circuit
shown in Fig. (a)
C2
12.0 µ F
C1
4.0 µ F
C3
3.0 µ F
B
C4
1.0 µ F
(a)
A
C2
12.0 µ F
C1
4.0 µ F
C34
4.0 µ F
B
(b)
If we apply a voltage of 100V
bet ween points A and B
what is the voltage across
capacitor C2?
A
C1
4.0 µ F
C234
3.0 µ F
B
(c)
A
7.0 µ F
Ans: 25V
B
(d)
1. (Ceq)a > (Ceq)b = (Ceq)c > (Ceq)d
Rank in order, from largest
to smallest, the equivalent
capacitance (Ceq)a to (Ceq)d
2. (Ceq)b > (Ceq)a = (Ceq)d > (Ceq)c
of circuits a to d.
4. (Ceq)d > (Ceq)b = (Ceq)c > (Ceq)a
3. (Ceq)c > (Ceq)a = (Ceq)d > (Ceq)b
5. (Ceq)d > (Ceq)b > (Ceq)a > (Ceq)c
1. (Ceq)a > (Ceq)b = (Ceq)c > (Ceq)d
Rank in order, from largest
to smallest, the equivalent
capacitance (Ceq)a to (Ceq)d
2. (Ceq)b > (Ceq)a = (Ceq)d > (Ceq)c
of circuits a to d.
4. (Ceq)d > (Ceq)b = (Ceq)c > (Ceq)a
3. (Ceq)c > (Ceq)a = (Ceq)d > (Ceq)b
5. (Ceq)d > (Ceq)b > (Ceq)a > (Ceq)c
C=5
C=3+3=6
C=1/(1/3+1/3)=3/2
C=3+1/(1/4+1/4)=5
1) Ceq = 3/2 C
What is the equivalent capacitance,
2) Ceq = 2/3 C
Ceq , of the combination below?
3) Ceq = 3 C
4) Ceq = 1/3 C
5) Ceq = 1/2 C
o
Ceq
o
C
C
C
1) Ceq = 3/2 C
What is the equivalent capacitance,
2) Ceq = 2/3 C
Ceq , of the combination below?
3) Ceq = 3 C
4) Ceq = 1/3 C
5) Ceq = 1/2 C
The 2 equal capacitors in series add
up as inverses, giving 1/2 C. These
are parallel to the first one, which
add up directly. Thus, the total
equivalent capacitance is 3/2 C.
o
Ceq
o
C
C
C
How does the voltage V1 across
1) V1 = V2
the first capacitor (C1) compare
2) V1 > V2
to the voltage V2 across the
3) V1 < V2
second capacitor (C2)?
4) all voltages are zero
C2 = 1.0 µF
10 V
C1 = 1.0 µF
C3 = 1.0 µF
How does the voltage V1 across
1) V1 = V2
the first capacitor (C1) compare
2) V1 > V2
to the voltage V2 across the
3) V1 < V2
second capacitor (C2)?
4) 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 µF
10 V
C1 = 1.0 µF
C3 = 1.0 µF
How does the charge Q1 on the first
capacitor (C1) compare to the
1) Q1 = Q2
2) Q1 > Q2
charge Q2 on the second capacitor
3) Q1 < Q2
(C2)?
4) all charges are zero
C2 = 1.0 µF
10 V
C1 = 1.0 µF
C3 = 1.0 µF
How does the charge Q1 on the first
capacitor (C1) compare to the
1) Q1 = Q2
2) Q1 > Q2
charge Q2 on the second capacitor
3) Q1 < Q2
(C2)?
4) all charges are zero
We already know that the
voltage across C1 is 10 V
C2 = 1.0 µF
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 µF
C3 = 1.0 µF
Initially the switch is in position A and capacitors C2 and C3 are uncharged. Then
the switch is flipped to position B. After ward, what are the charge on and the
potential difference across each capacitor?
Ans: Q1 = 0.83 mC, Q2 = Q3 = 0.63 mC
V1 = 55 V, V2 = 34 V, V3 = 21 V
You reposition the two plates of a
capacitor so that the capacitance
doubles.
1. increases by a factor of 4
If the charges +Q and –Q on the two
plates are kept constant in this
process, what happens to the energy
stored in the capacitor?
3. the stored energy is unchanged
2. increases by a factor of 2
4. decreases by a factor of 2
5. decreases by a factor of 4
You reposition the two plates of a
capacitor so that the capacitance
doubles.
1. increases by a factor of 4
If the charges +Q and –Q on the two
plates are kept constant in this
process, what happens to the energy
stored in the capacitor?
3. the stored energy is unchanged
2. increases by a factor of 2
4. decreases by a factor of 2
5. decreases by a factor of 4
You slide a slab of dielectric
between the plates of a
capacitor. As you do this, the
charges on the plates remain
constant.
What effect does this have
on the energy stored in the
capacitor?
+Q –Q
1. the stored energy increases
2. the stored energy is unchanged
3. the stored energy decreases
4. not enough information given to decide
+Q
–Q
You slide a slab of dielectric
between the plates of a
capacitor. As you do this, the
charges on the plates remain
constant.
What effect does this have
on the energy stored in the
capacitor?
+Q –Q
1. the stored energy increases
2. the stored energy is unchanged
3. the stored energy decreases
4. not enough information given to decide
+Q
–Q
Two 5.0-cm-diameter metal disks separated by a 0.50-mm-thick piece of Pyrex glass
are charged to a potential difference of 1000 V. What are
(a) the surface charge density on the disks, and
(b) the surface charge density on the glass?
Look up in table 30.1, for Pyrex glass κ = 4.7
Ans: (a) 8.3 x 10-5 C/m2
(b) 1.4 x 10-5 C/m2