Capacitors
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Basic Formulas for capacitors:
Definition of capacitance: C 
Q
V
The standard unit for C is the Farad.
Formula relating voltage across plates to the electric field strength for a parallelplate capacitor:
V  E  d
Diagram of a parallel-plate capacitor d
+ + + + + + + + + + + +
E
_ _ _ _ _ _ _ _ _ _ _ _ _
+
_
Voltage
Source
Energy stored in a capacitor:
U elec
 1
2
CV 2  1
2
QV  1
2
Q 2
C
Capacitors in Parallel:
C
1
C eq
 C
1
 C
2
Capacitors in Series:
C
1
C
2
1
C eq
C eq


1
C
1

C
1
C
1


C
2
C
2
1
C
2
Shortcut – works for any pair of capacitors in series.
C
2
Voltage across C
1 and C
2 must be equal.
Charge on each may be different.
Voltage across C
1 and C
2 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
_
_ _ _ _ _ _ _ _ _ _ _ _ _
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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.
+ + + + + + + + + + + +
+
0.75 cm +
12 V
_
_ _ _ _ _ _ _ _ _ _ _ _ _
V = E∙d
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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.
+ + + + + + + + + + + +
+
0.75 cm
_ _ _ _ _ _ _ _ _ _ _ _ _
V = E∙d
12 V  E  ( 0 .
0075 m )  E  1600 V m
Bonus Question: Which direction does the E-field point?
+
12 V
_
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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.
+ + + + + + + + + + + +
+
0.75 cm
E
_ _ _ _ _ _ _ _ _ _ _ _ _
V = E∙d
12 V  E  ( 0 .
0075 m )  E  1600 V m
Bonus Question: Which direction does the E-field point?
Downward (away from + charge and toward - )
+
12 V
_
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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.
+ + + + + + + + + + + +
+
0.75 cm
E
_ _ _ _ _ _ _ _ _ _ _ _ _
V = E∙d
12 V  E  ( 0 .
0075 m )  E  1600 V m
Bonus Question: Which direction does the E-field point?
Downward (away from + charge and toward - )
+
12 V
_
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.
+ + + + + + + + + + + +
+
0.75 cm
E
_ _ _ _ _ _ _ _ _ _ _ _ _
V = E∙d
12 V  E  ( 0 .
0075 m )  E  1600 V m
Bonus Question: Which direction does the E-field point?
Downward (away from + charge and toward - )
+
12 V
_
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:
 U elec
 q   V  ( 6 .
24  10  6 C )(  12 V )   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 m

7 .
49
1
2 m  v 2
 10  5 J  1
2 m  ( 3 .
4 m s
) 2
 1 .
3  10  5 kg
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C
1
=6μF; C
2
=2μF
C
1
6V
C
2
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
C
2
6V
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
C
2
C eq
6V
6V
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
C
2
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
Q  CV  (
3
2
 F )( 6 V )  9  C
C eq
6V
6V
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
C
2
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
Q  CV  (
3
2
 F )( 6 V )  9  C
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).
C eq
6V
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.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
C
2
C eq
Q  CV  (
3
2
 F )( 6 V )  9  C
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).
Q
1
 Q
2
 9  C
6V
6V
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB

Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
C
2
C eq
Q  CV  (
3
2
 F )( 6 V )  9  C
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).
Q
1
 Q
2
 9  C
Rearranging our basic formula and applying it to each individual capacitor gives us the voltage across each:
6V
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.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
C
2
C eq
Q  CV  (
3
2
 F )( 6 V )  9  C
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).
Q
1
 Q
2
 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.
V
1
V
2


Q
1
C
1
Q
2
C
2

9  C
6  F

9  C
2  F

3
2
Volts

9
2
Volts
6V
6V
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Circuit Example #1:
Find the voltage across, and energy stored in each capacitor in the circuit shown.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
C
2
C eq
Q  CV  (
3
2
 F )( 6 V )  9  C
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).
Q
1
 Q
2
 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.
V
1
V
2


Q
1
C
1
Q
2
C
2

9  C
6  F

9  C
2  F

3
2
Volts

9
2
Volts
Our final calculations use a formula for stored energy:
U elec , 1
U elec , 2


1
2
Q
1
V
1
1
2
Q
2
V
2


1
2
( 9  C )(
3
2
1
2
( 9  C )(
9
2
V )
V )


27
4
81
4
 J
 J
6V
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.
C
1
=6μF; C
2
=2μF
C
1
These capacitors are in series.
Use the formula to find the equivalent capacitance:
C eq

6  F
6  F


2  F
2  F

12
8
 F 
3
2
 F
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
C
2
C eq
Q  CV  (
3
2
 F )( 6 V )  9  C
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).
Q
1
 Q
2
 9  C
6V
6V
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.
V
1
V
2


Q
1
C
1
Q
2
C
2

9  C
6  F

9  C
2  F

3
2
Volts

9
2
Volts
Our final calculations use a formula for stored energy:
U elec , 1
U elec , 2


1
2
Q
1
V
1
1
2
Q
2
V
2


1
2
( 9  C )(
3
2
1
2
( 9  C )(
9
2
V )
V )


27
4
81
4
 J
 J
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.
C
1
=1μF; C
2
=2μF; C
3
=3μF
C
1
C
2
C
3
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C
1
=1μF; C
2
=2μF; C
3
=3μF
We need to find the equivalent capacitance for this circuit, then work backwards to find the energy in each capacitor.
C
1
C
2
C
3
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C
1
=1μF; C
2
=2μF; C
3
=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 C capacitance C
1
+C
2
1 and C
2 are in parallel to each other, so they are equivalent to a single capacitor with
=3µF. Draw a new diagram for this:
C
1
C
2
C
3
C
1
+C
2
C
3
6V
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C
1
=1μF; C
2
=2μF; C
3
=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 C capacitance C
1
+C
2
1 and C
2 are in parallel to each other, so they are equivalent to a single capacitor with
=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:
C eq

3

F

3

F
3

F

3

F

9
6

F

3
2

F
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C
1
=1μF; C
2
=2μF; C
3
=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 C capacitance C
1
+C
2
1 and C
2 are in parallel to each other, so they are equivalent to a single capacitor with
=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:
C eq

3

F

3

F
3

F

3

F

9
6

F

3
2

F
The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:
Q  CV  (
3
2
 F )( 6 V )  9  C
Next we will work backwards to find the information about each individual capacitor:
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C
1
=1μF; C
2
=2μF; C
3
=3μF
It may help to set up a table like this to keep track of all the info.
Charge Energy
C
C
C
C
1
2
3 eq
Capac.
Voltage
1µF
2µF
3µF
1.5µF 6V 9µC
This is what we know so far.
The next step is to realize that the charge on C
3 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 C
3
.
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C
1
=1μF; C
2
=2μF; C
3
=3μF
It may help to set up a table like this to keep track of all the info.
Charge Energy
C
C
C
C
1
2
3 eq
Capac.
Voltage
1µF
2µF
3µF
1.5µF 6V
9µC
9µC
This is what we know so far.
The next step is to realize that the charge on C
3 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 C
3
.
So we can fill in the charge on C
3
.
Now that we have the charge we can find the voltage as well:
V
3

Q
3
C
3

9  C
3  F
 3 Volts
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
6V
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Circuit Example #2:
Find the energy stored in each capacitor in the circuit shown.
C
1
=1μF; C
2
=2μF; C
3
=3μF
It may help to set up a table like this to keep track of all the info.
Charge Energy
C
C
C
C
1
2
3 eq
Capac.
Voltage
1µF
2µF
3µF 3V
1.5µF 6V
9µC
9µC
This is what we know so far.
The next step is to realize that the charge on C
3 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 C
3
.
So we can fill in the charge on C
3
.
Now that we have the charge we can find the voltage as well:
V
3

Q
3
C
3

9
3


C
F
 3 Volts
We can also find the energy stored in C
3
, as well as the total.
U
3

U total
1
2

Q
3
1
2
V
3

Q total
V
1
2
( 9  C total

)(
1
2
3 V ) 
27
2
( 9  C )( 6 V )
 J
 27  J
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
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.
C
1
=1μF; C
2
=2μF; C
3
=3μF
It may help to set up a table like this to keep track of all the info.
Charge Energy
C
C
C
C
1
2
3 eq
Capac.
Voltage
1µF
2µF
3µF 3V
1.5µF 6V
9µC
9µC
13.5µJ
27µJ
This is what we know so far.
Next we have to figure out the info for C
1 and C
2
.
These are parallel capacitors, so they should have the same voltage.
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
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.
C
1
=1μF; C
2
=2μF; C
3
=3μF
It may help to set up a table like this to keep track of all the info.
Charge Energy
C
C
C
C
1
2
3 eq
Capac.
Voltage
1µF
2µF
3µF 3V
1.5µF 6V
9µC
9µC
13.5µJ
27µJ
This is what we know so far.
Next we have to figure out the info for C
1 and C
2
.
These are parallel capacitors, so they should have the same voltage.
We know the total voltage is 6V, and since the voltage on C and C
2
3 with the others) is 3V, that leaves 3V left for C
1 is that the voltages have to add up when you make a complete loop around the circuit.
(in series
. The basic rule
So let’s fill in those boxes in the table:
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
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.
C
1
=1μF; C
2
=2μF; C
3
=3μF
C
1
C
2
It may help to set up a table like this to keep track of all the info.
Charge Energy
C
C
C
C
1
2
3 eq
Capac.
Voltage
1µF
2µF
3V
3V
3µF 3V
1.5µF 6V
9µC
9µC
13.5µJ
27µJ
C
3
C
1
+C
2
This is what we know so far.
Next we have to figure out the info for C
1 and C
2
.
These are parallel capacitors, so they should have the same voltage.
We know the total voltage is 6V, and since the voltage on C and C
2
3 with the others) is 3V, that leaves 3V left for C
1 is that the voltages have to add up when you make a complete loop around the circuit.
(in series
. The basic rule
So let’s fill in those boxes in the table:
C eq
For completeness let’s find the charge on C
1
Q
Q
1
2


C
C
1
2
V
V
1
2


(
(
1
2


F
F
)( 3
)( 3 v v
)
)


3
6

 c c and C
2 as well:
Notice that the charge adds up to the total, as it should.
C
3
6V
6V
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.
C
1
=1μF; C
2
=2μF; C
3
=3μF
It may help to set up a table like this to keep track of all the info.
Energy
C
C
C
C
1
2
3 eq
Capac.
Voltage
1µF
2µF
3V
3V
3µF 3V
1.5µF 6V
Charge
3µC
6µC
9µC
9µC
13.5µJ
27µJ
Finally we can calculate the energy stored in C
1
U
U
1
2


1
2
1
2
Q
1
V
1
Q
2
V
2


1
2
1
2
( 3  C )( 3 V )
( 6  C )( 3 V )


9
2
 J
9  J and C
2
, and we are done.
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
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.
C
1
=1μF; C
2
=2μF; C
3
=3μF
It may help to set up a table like this to keep track of all the info.
C
C
C
C
1
2
3 eq
Capac.
Voltage
1µF
2µF
3V
3V
3µF 3V
1.5µF 6V
Charge
3µC
6µC
9µC
9µC
Energy
4.5µJ
9µJ
13.5µJ
27µJ
Finally we can calculate the energy stored in C
1
U
U
1
2


1
2
1
2
Q
1
V
1
Q
2
V
2


1
2
1
2
( 3  C )( 3 V )
( 6  C )( 3 V )


9
2
 J
9  J and C
2
, and we are done.
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.
C
1
C
2
C
3
C
1
+C
2
C eq
C
3
6V
6V
6V
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.
C with
=κ·C without
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