Transcript Physics 6B - University of California, Santa Barbara

Physics 6B
Capacitors
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Basic Formulas for capacitors:
Q
V
The standard unit for C is the Farad.
Diagram of a parallel-plate capacitor
Definition of capacitance: C 
Formula relating voltage across plates to
the electric field strength for a parallelplate capacitor: V  E  d
+
+
+
+
+
+
+
+
+
+
+
+
E
d
+
_
Voltage
Source
_
_
_
_
_
_
_
_
_
_
_
_
_
Energy stored in a capacitor:
2
Uelec  CV  QV 
1
2
1
2
1
2
Q2
C
Capacitors in Parallel:
C1
Ceq  C1  C2
Capacitors in Series:
1
1
1


Ceq C1 C2
C1
C2
Ceq 
C1  C2
C1  C2
Shortcut – works for any
pair of capacitors in series.
C2
Voltage across C1 and C2 must be equal.
Charge on each may be different.
Voltage across C1 and C2 may be different.
Charge on each must be equal.
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Example: A uniform electric field is established by connecting the plates of a parallel-plate
capacitor to a 12-volt battery. (a) If the plates are separated by 0.75cm, what is the
magnitude of the electric field in the capacitor? (b) A charge of +6.24 x 10-6 C moves from the
positive plate to the negative plate. Find the change in electric potential energy for this charge.
+
+
+
+
+
+
+
+
+
+
+
+
+
0.75 cm
+
_
12 V
_
_
_
_
_
_
_
_
_
_
_
_
_
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example: A uniform electric field is established by connecting the plates of a parallel-plate
capacitor to a 12-volt battery. (a) If the plates are separated by 0.75cm, what is the
magnitude of the electric field in the capacitor? (b) A charge of +6.24 x 10-6 C moves from the
positive plate to the negative plate. Find the change in electric potential energy for this charge.
For a parallel-plate capacitor we have a
very simple formula relating the voltage to
the electric field inside.
V = E∙d
+
+
+
+
+
+
+
+
+
+
+
+
+
0.75 cm
+
_
12 V
_
_
_
_
_
_
_
_
_
_
_
_
_
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example: A uniform electric field is established by connecting the plates of a parallel-plate
capacitor to a 12-volt battery. (a) If the plates are separated by 0.75cm, what is the
magnitude of the electric field in the capacitor? (b) A charge of +6.24 x 10-6 C moves from the
positive plate to the negative plate. Find the change in electric potential energy for this charge.
For a parallel-plate capacitor we have a
very simple formula relating the voltage to
the electric field inside.
V = E∙d
+
+
+
+
+
+
+
+
+
+
+
+
+
0.75 cm
+
_
12 V
_
_
_
_
_
_
_
_
_
_
_
_
_
V
12V  E  (0.0075m)  E  1600 m
Bonus Question: Which direction does the E-field point?
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Example: A uniform electric field is established by connecting the plates of a parallel-plate
capacitor to a 12-volt battery. (a) If the plates are separated by 0.75cm, what is the
magnitude of the electric field in the capacitor? (b) A charge of +6.24 x 10-6 C moves from the
positive plate to the negative plate. Find the change in electric potential energy for this charge.
For a parallel-plate capacitor we have a
very simple formula relating the voltage to
the electric field inside.
V = E∙d
+
+
+
+
+
+
+
+
+
+
+
+
+
E
0.75 cm
+
_
12 V
_
_
_
_
_
_
_
_
_
_
_
_
_
V
12V  E  (0.0075m)  E  1600 m
Bonus Question: Which direction does the E-field point?
Downward (away from + charge and toward - )
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Example: A uniform electric field is established by connecting the plates of a parallel-plate
capacitor to a 12-volt battery. (a) If the plates are separated by 0.75cm, what is the
magnitude of the electric field in the capacitor? (b) A charge of +6.24 x 10-6 C moves from the
positive plate to the negative plate. Find the change in electric potential energy for this charge.
For a parallel-plate capacitor we have a
very simple formula relating the voltage to
the electric field inside.
V = E∙d
+
+
+
+
+
+
+
+
+
+
+
+
+
E
0.75 cm
+
_
12 V
_
_
_
_
_
_
_
_
_
_
_
_
_
V
12V  E  (0.0075m)  E  1600 m
Bonus Question: Which direction does the E-field point?
Downward (away from + charge and toward - )
For part (b) we need to remember what exactly voltage means.
Each volt of potential difference represents 1 Joule of energy for each
Coulomb of charge. So if we multiply the voltage and the charge, we
get the change in the energy.
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Example: A uniform electric field is established by connecting the plates of a parallel-plate
capacitor to a 12-volt battery. (a) If the plates are separated by 0.75cm, what is the
magnitude of the electric field in the capacitor? (b) A charge of +6.24 x 10-6 C moves from the
positive plate to the negative plate. Find the change in electric potential energy for this charge.
For a parallel-plate capacitor we have a
very simple formula relating the voltage to
the electric field inside.
+
+
+
+
+
+
+
+
+
+
+
+
E
0.75 cm
V = E∙d
+
+
_
12 V
_
_
_
_
_
_
_
_
_
_
_
_
_
V
12V  E  (0.0075m)  E  1600 m
Bonus Question: Which direction does the E-field point?
Downward (away from + charge and toward - )
For part (b) we need to remember what exactly voltage means.
Each volt of potential difference represents 1 Joule of energy for each
Coulomb of charge. So if we multiply the voltage and the charge, we
get the change in the energy. So our answer is:
Uelec  q  V  (6.24  106 C)(12V)  7.49  105 J
Note that the answer is negative in this case. This is because we have a
postive charge moving with the E-field. As a general rule, if the charge is
moving in the direction that you expect the E-field to push it, then it is
losing potential energy and gaining kinetic energy.
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Suppose the charge from the previous problem is released from rest at the
positive plate and that it reaches the negative plate with speed 3.4 m/s.
What is the mass of the charge and its final kinetic energy?
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Suppose the charge from the previous problem is released from rest at the
positive plate and that it reaches the negative plate with speed 3.4 m/s.
What is the mass of the charge and its final kinetic energy?
For this one, just remember that when the positive charge is moving with
the field, it is picking up kinetic energy as it loses potential energy.
We just calculated the amount in the previous problem.
K  7.49  105 J
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Suppose the charge from the previous problem is released from rest at the
positive plate and that it reaches the negative plate with speed 3.4 m/s.
What is the mass of the charge and its final kinetic energy?
For this one, just remember that when the positive charge is moving with
the field, it is picking up kinetic energy as it loses potential energy.
We just calculated the amount in the previous problem.
K  7.49  105 J
We can now calculate the mass from our definition of kinetic energy.
K  12 m  v2
7.49  105 J  12 m  (3.4 ms )2
m  1.3  105kg
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
6V
C2
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6F  2F 12
3


F  F
6F  2F
8
2
6V
C2
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6F  2F 12
3


F  F
6F  2F
8
2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
6V
C2
Ceq
6V
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6F  2F 12
3


F  F
6F  2F
8
2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
3
Q  CV  ( F)(6V)  9C
2
6V
C2
Ceq
6V
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Assistance Services at UCSB
Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6F  2F 12
3


F  F
6F  2F
8
2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
3
Q  CV  ( F)(6V)  9C
2
6V
C2
Ceq
6V
This is the charge on the (fictional) equivalent capacitor. However, by looking
at the original diagram we see that the charge on each of the series capacitors
must be equal to this total (there is nowhere else for the charges to go).
Prepared by Vince Zaccone
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Assistance Services at UCSB
Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6F  2F 12
3


F  F
6F  2F
8
2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
3
Q  CV  ( F)(6V)  9C
2
6V
C2
Ceq
6V
This is the charge on the (fictional) equivalent capacitor. However, by looking
at the original diagram we see that the charge on each of the series capacitors
must be equal to this total (there is nowhere else for the charges to go).
Q1  Q2  9C
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Assistance Services at UCSB
Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6F  2F 12
3


F  F
6F  2F
8
2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
3
Q  CV  ( F)(6V)  9C
2
6V
C2
Ceq
6V
This is the charge on the (fictional) equivalent capacitor. However, by looking
at the original diagram we see that the charge on each of the series capacitors
must be equal to this total (there is nowhere else for the charges to go).
Q1  Q2  9C
Rearranging our basic formula and applying it to each individual
capacitor gives us the voltage across each:
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6V
6F  2F 12
3


F  F
6F  2F
8
2
C2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
Ceq
3
Q  CV  ( F)(6V)  9C
2
6V
This is the charge on the (fictional) equivalent capacitor. However, by looking
at the original diagram we see that the charge on each of the series capacitors
must be equal to this total (there is nowhere else for the charges to go).
Q1  Q2  9C
Rearranging our basic formula and applying it to each individual
capacitor gives us the voltage across each:
Notice that the total voltage adds up to 6V, as it should.
V1 
Q1 9C 3

 Volts
C1 6F 2
V2 
Q2 9C 9

 Volts
C2
2F 2
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6V
6F  2F 12
3


F  F
6F  2F
8
2
C2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
Ceq
3
Q  CV  ( F)(6V)  9C
2
6V
This is the charge on the (fictional) equivalent capacitor. However, by looking
at the original diagram we see that the charge on each of the series capacitors
must be equal to this total (there is nowhere else for the charges to go).
Q1  Q2  9C
Rearranging our basic formula and applying it to each individual
capacitor gives us the voltage across each:
Notice that the total voltage adds up to 6V, as it should.
V1 
Q1 9C 3

 Volts
C1 6F 2
Our final calculations use a formula for stored energy:
V2 
Q2 9C 9

 Volts
C2
2F 2
3
27
Uelec ,1  12 Q1V1  12 (9C)( V) 
J
2
4
9
81
Uelec ,2  12 Q2 V2  12 (9C)( V) 
J
2
4
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C1=6μF; C2=2μF
C1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
Ceq
6V
6F  2F 12
3


F  F
6F  2F
8
2
C2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
Ceq
3
Q  CV  ( F)(6V)  9C
2
6V
This is the charge on the (fictional) equivalent capacitor. However, by looking
at the original diagram we see that the charge on each of the series capacitors
must be equal to this total (there is nowhere else for the charges to go).
Q1  Q2  9C
Rearranging our basic formula and applying it to each individual
capacitor gives us the voltage across each:
Notice that the total voltage adds up to 6V, as it should.
V1 
Q1 9C 3

 Volts
C1 6F 2
Our final calculations use a formula for stored energy:
V2 
Q2 9C 9

 Volts
C2
2F 2
3
27
Uelec ,1  12 Q1V1  12 (9C)( V) 
J
2
4
9
81
Uelec ,2  12 Q2 V2  12 (9C)( V) 
J
2
4
Note that the total energy adds up to 27μJ. This is what we would get if we used
the single equivalent capacitance of 1.5 μF and the total battery voltage of 6V.
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
6V
C3
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
We need to find the equivalent capacitance for this circuit,
then work backwards to find the energy in each capacitor.
C1
C2
6V
C3
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
We need to find the equivalent capacitance for this circuit,
then work backwards to find the energy in each capacitor.
The first step is to recognize that C1 and C2 are in parallel to
each other, so they are equivalent to a single capacitor with
capacitance C1+C2=3µF. Draw a new diagram for this:
C1
C2
6V
C3
C1+C2
6V
C3
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
We need to find the equivalent capacitance for this circuit,
then work backwards to find the energy in each capacitor.
The first step is to recognize that C1 and C2 are in parallel to
each other, so they are equivalent to a single capacitor with
capacitance C1+C2=3µF. Draw a new diagram for this:
Now we see that the remaining capacitors are in series, so
we use the reciprocal formula to find the equivalent
capacitance. Draw a new diagram:
Ceq 
3F  3F 9
3
 F  F
3F  3F 6
2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
C1
C2
6V
C3
C1+C2
6V
C3
Ceq
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
We need to find the equivalent capacitance for this circuit,
then work backwards to find the energy in each capacitor.
The first step is to recognize that C1 and C2 are in parallel to
each other, so they are equivalent to a single capacitor with
capacitance C1+C2=3µF. Draw a new diagram for this:
Now we see that the remaining capacitors are in series, so
we use the reciprocal formula to find the equivalent
capacitance. Draw a new diagram:
Ceq 
3F  3F 9
3
 F  F
3F  3F 6
2
The new diagram has just a single capacitor. Now we can
use the definition of capacitance to find the charge:
C1
C2
6V
C3
C1+C2
6V
C3
Ceq
6V
3
Q  CV  ( F)(6V)  9C
2
Next we will work backwards to find the information about
each individual capacitor:
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
It may help to set up a table like this to keep track of all the info.
Capac.
C1
1µF
C2
2µF
C3
3µF
Ceq
1.5µF
Voltage
Charge
Energy
6V
C3
C1+C2
6V
6V
9µC
C3
This is what we know so far.
The next step is to realize that the charge on C3 must be the total
charge. Take a look at the middle diagram (or the original one) and
convince yourself that all the charge must land on C3.
Ceq
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
It may help to set up a table like this to keep track of all the info.
Capac.
C1
1µF
C2
2µF
C3
3µF
Ceq
1.5µF
Voltage
Charge
Energy
6V
C3
C1+C2
9µC
6V
6V
9µC
C3
This is what we know so far.
The next step is to realize that the charge on C3 must be the total
charge. Take a look at the middle diagram (or the original one) and
convince yourself that all the charge must land on C3.
Ceq
6V
So we can fill in the charge on C3.
Now that we have the charge we can find the voltage as well:
V3 
Q3 9C

 3Volts
C3
3F
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
It may help to set up a table like this to keep track of all the info.
Capac.
Voltage
Charge
C1
1µF
C2
2µF
C3
3µF
3V
9µC
Ceq
1.5µF
6V
9µC
Energy
6V
C3
C1+C2
6V
C3
This is what we know so far.
The next step is to realize that the charge on C3 must be the total
charge. Take a look at the middle diagram (or the original one) and
convince yourself that all the charge must land on C3.
Ceq
6V
So we can fill in the charge on C3.
Now that we have the charge we can find the voltage as well:
V3 
Q3 9C

 3Volts
C3
3F
We can also find the energy stored in C3, as well as the total.
27
J
2
 12 (9C)(6V)  27J
U3  12 Q3V3  12 (9C)(3V) 
Utotal  12 QtotalVtotal
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
It may help to set up a table like this to keep track of all the info.
Capac.
Voltage
Charge
Energy
C1
1µF
C2
2µF
C3
3µF
3V
9µC
13.5µJ
Ceq
1.5µF
6V
9µC
27µJ
6V
C3
C1+C2
6V
C3
This is what we know so far.
Next we have to figure out the info for C1 and C2.
These are parallel capacitors, so they should have the same voltage.
Ceq
6V
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
It may help to set up a table like this to keep track of all the info.
Capac.
Voltage
Charge
Energy
C1
1µF
C2
2µF
C3
3µF
3V
9µC
13.5µJ
Ceq
1.5µF
6V
9µC
27µJ
6V
C3
C1+C2
6V
C3
This is what we know so far.
Next we have to figure out the info for C1 and C2.
These are parallel capacitors, so they should have the same voltage.
Ceq
6V
We know the total voltage is 6V, and since the voltage on C3 (in series
with the others) is 3V, that leaves 3V left for C1 and C2. The basic rule
is that the voltages have to add up when you make a complete loop
around the circuit.
So let’s fill in those boxes in the table:
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
It may help to set up a table like this to keep track of all the info.
Capac.
Voltage
Charge
Energy
C1
1µF
3V
C2
2µF
3V
C3
3µF
3V
9µC
13.5µJ
Ceq
1.5µF
6V
9µC
27µJ
6V
C3
C1+C2
6V
C3
This is what we know so far.
Ceq
Next we have to figure out the info for C1 and C2.
These are parallel capacitors, so they should have the same voltage.
6V
We know the total voltage is 6V, and since the voltage on C3 (in series
with the others) is 3V, that leaves 3V left for C1 and C2. The basic rule
is that the voltages have to add up when you make a complete loop
around the circuit.
So let’s fill in those boxes in the table:
For completeness let’s find the charge on C1 and C2 as well:
Q1  C1V1  (1F)(3v)  3c
Q2  C2V2  (2F)(3v)  6c
Notice that the charge adds up to the total, as it should.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
It may help to set up a table like this to keep track of all the info.
Capac.
Voltage
Charge
Energy
C1
1µF
3V
3µC
C2
2µF
3V
6µC
C3
3µF
3V
9µC
13.5µJ
Ceq
1.5µF
6V
9µC
27µJ
6V
C3
C1+C2
6V
C3
Finally we can calculate the energy stored in C1 and C2, and we are done.
9
J
2
U2  12 Q2V2  12 (6C)(3V)  9J
Ceq
U1  12 Q1V1  12 (3C)(3V) 
6V
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C1=1μF; C2=2μF; C3=3μF
C1
C2
It may help to set up a table like this to keep track of all the info.
Capac.
Voltage
Charge
Energy
C1
1µF
3V
3µC
4.5µJ
C2
2µF
3V
6µC
9µJ
C3
3µF
3V
9µC
13.5µJ
Ceq
1.5µF
6V
9µC
27µJ
6V
C3
C1+C2
6V
C3
Finally we can calculate the energy stored in C1 and C2, and we are done.
9
J
2
U2  12 Q2V2  12 (6C)(3V)  9J
Ceq
U1  12 Q1V1  12 (3C)(3V) 
6V
Note that we can check our answers to make sure they add up. The total
energy provided by the battery should match up with the sum of the
energies of the 3 individual capacitors, and it does.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Dielectrics:
When an insulating material is inserted
between the plates of a capacitor, its
capacitance increases.
Cwith=κ·Cwithout
κ is called the Dielectric Constant
The presence of a dielectric weakens the
net electric field between the plates,
allowing more charge to build up (thus
increasing the capacity to hold charge)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB