Physics 6B

Capacitors

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Basic Formulas for capacitors:

Definition of capacitance:

The standard unit for C is the Farad.

Formula relating voltage across plates to the electric field strength for a parallel-plate capacitor:

Energy stored in a capacitor:

VQ

C

Voltage Source

+ _

+ + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _

dE

Diagram of a parallel-plate capacitor

dEV

CQ

QVCVU2

21

212

21

elec

Capacitors in Parallel:

Voltage across C1 and C2 must be equal.Charge on each may be different.

C1

C2

21eq CCC

Capacitors in Series:

Voltage across C1 and C2 may be different.Charge on each must be equal.

C1 C2

21

21eq

21eq

CCCC

C

C1

C1

C1

Shortcut – works for any pair of capacitors in series.

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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.

12 V

+ _

+ + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _

0.75 cm

+

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.

12 V

+ _

+ + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _

0.75 cm

+For a parallel-plate capacitor we have a very simple formula relating the voltage to the electric field inside.

V = E∙d

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.

12 V

+ _

+ + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _

0.75 cm

+For a parallel-plate capacitor we have a very simple formula relating the voltage to the electric field inside.

V = E∙d

mV1600E)m0075.0(EV12

Bonus Question: Which direction does the E-field point?

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.

12 V

+ _

+ + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _

0.75 cm

+For a parallel-plate capacitor we have a very simple formula relating the voltage to the electric field inside.

V = E∙d

mV1600E)m0075.0(EV12

Bonus Question: Which direction does the E-field point? Downward (away from + charge and toward - )

E

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.

12 V

+ _

+ + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _

0.75 cm

+For a parallel-plate capacitor we have a very simple formula relating the voltage to the electric field inside.

V = E∙d

mV1600E)m0075.0(EV12

Bonus Question: Which direction does the E-field point? Downward (away from + charge and toward - )

E

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.

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.

12 V

+ _

+ + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _ _

0.75 cm

+For a parallel-plate capacitor we have a very simple formula relating the voltage to the electric field inside.

V = E∙d

mV1600E)m0075.0(EV12

Bonus Question: Which direction does the E-field point? Downward (away from + charge and toward - )

E

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:

J1049.7)V12)(C1024.6(VqU 56elec

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.

J1049.7K 5

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For Campus Learning Assistance Services at UCSB

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.

J1049.7K 5

We can now calculate the mass from our definition of kinetic energy.

kg103.1m

)4.3(mJ1049.7

vmK

5

2sm

215

221

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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

C2

C1

6V

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C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

C9)V6)(F23

(CVQ

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

C9)V6)(F23

(CVQ

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).

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

C9)V6)(F23

(CVQ

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).

C9QQ 21

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

C9)V6)(F23

(CVQ

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).

C9QQ 21

Rearranging our basic formula and applying it to each individual capacitor gives us the voltage across each:

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

C9)V6)(F23

(CVQ

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).

C9QQ 21

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.

Volts29

F2C9

CQ

V

Volts23

F6C9

CQ

V

2

22

1

11

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

C9)V6)(F23

(CVQ

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).

C9QQ 21

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.

Volts29

F2C9

CQ

V

Volts23

F6C9

CQ

V

2

22

1

11

Our final calculations use a formula for stored energy:

J481

)V29

)(C9(VQU

J4

27)V

23

)(C9(VQU

21

2221

2,elec

21

1121

1,elec

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C2

C1

6V

These capacitors are in series.Use the formula to find the equivalent capacitance:

F23

F8

12F2F6F2F6

Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

C9)V6)(F23

(CVQ

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).

C9QQ 21

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.

Volts29

F2C9

CQ

V

Volts23

F6C9

CQ

V

2

22

1

11

Our final calculations use a formula for stored energy:

J481

)V29

)(C9(VQU

J4

27)V

23

)(C9(VQU

21

2221

2,elec

21

1121

1,elec

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.

Circuit Example #1:Find the voltage across, and energy stored in each capacitor in the circuit shown.

C1=6μF; C2=2μF

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

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

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For Campus Learning Assistance Services at UCSB

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

We need to find the equivalent capacitance for this circuit, then work backwards to find the energy in each capacitor.

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

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

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:

C3

C1+C2

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

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

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:

C3

C1+C2

6VNow we see that the remaining capacitors are in series, so we use the reciprocal formula to find the equivalent capacitance. Draw a new diagram:

F2

3F

6

9

F3F3

F3F3Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

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

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

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:

C3

C1+C2

6VNow we see that the remaining capacitors are in series, so we use the reciprocal formula to find the equivalent capacitance. Draw a new diagram:

F2

3F

6

9

F3F3

F3F3Ceq

6V

Ceq

The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge:

C9)V6)(F23

(CVQ

Next we will work backwards to find the information about each individual capacitor:

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C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

C3

C1+C2

6V

6V

Ceq

Capac. Voltage Charge Energy

C1 1µF

C2 2µF

C3 3µF

Ceq 1.5µF 6V 9µC

It may help to set up a table like this to keep track of all the info.

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.

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

C3

C1+C2

6V

6V

Ceq

Capac. Voltage Charge Energy

C1 1µF

C2 2µF

C3 3µF 9µC

Ceq 1.5µF 6V 9µC

It may help to set up a table like this to keep track of all the info.

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.

So we can fill in the charge on C3.Now that we have the charge we can find the voltage as well:

Volts3F3C9

CQ

V3

33

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

C3

C1+C2

6V

6V

Ceq

Capac. Voltage Charge Energy

C1 1µF

C2 2µF

C3 3µF 3V 9µC

Ceq 1.5µF 6V 9µC

It may help to set up a table like this to keep track of all the info.

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.

So we can fill in the charge on C3.Now that we have the charge we can find the voltage as well:

Volts3F3C9

CQ

V3

33

We can also find the energy stored in C3, as well as the total.

J27)V6)(C9(VQU

J2

27)V3)(C9(VQU

21

totaltotal21

total

21

3321

3

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

C3

C1+C2

6V

6V

Ceq

Capac. Voltage Charge Energy

C1 1µF

C2 2µF

C3 3µF 3V 9µC 13.5µJ

Ceq 1.5µF 6V 9µC 27µJ

It may help to set up a table like this to keep track of all the info.

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.

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

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For Campus Learning Assistance Services at UCSB

C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

C3

C1+C2

6V

6V

Ceq

Capac. Voltage Charge Energy

C1 1µF

C2 2µF

C3 3µF 3V 9µC 13.5µJ

Ceq 1.5µF 6V 9µC 27µJ

It may help to set up a table like this to keep track of all the info.

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.

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:

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

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C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

C3

C1+C2

6V

6V

Ceq

Capac. Voltage Charge Energy

C1 1µF 3V

C2 2µF 3V

C3 3µF 3V 9µC 13.5µJ

Ceq 1.5µF 6V 9µC 27µJ

It may help to set up a table like this to keep track of all the info.

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.

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:

c6)v3)(F2(VCQ

c3)v3)(F1(VCQ

222

111

Notice that the charge adds up to the total, as it should.

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

C3

C1+C2

6V

6V

Ceq

Capac. Voltage Charge Energy

C1 1µF 3V 3µC

C2 2µF 3V 6µC

C3 3µF 3V 9µC 13.5µJ

Ceq 1.5µF 6V 9µC 27µJ

It may help to set up a table like this to keep track of all the info.

Finally we can calculate the energy stored in C1 and C2, and we are done.

J9)V3)(C6(VQU

J29

)V3)(C3(VQU

21

2221

2

21

1121

1

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

C1

C1=1μF; C2=2μF; C3=3μFC2

C3

6V

C3

C1+C2

6V

6V

Ceq

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

It may help to set up a table like this to keep track of all the info.

Finally we can calculate the energy stored in C1 and C2, and we are done.

J9)V3)(C6(VQU

J29

)V3)(C3(VQU

21

2221

2

21

1121

1

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.

Circuit Example #2:Find the energy stored in each capacitor in the circuit shown.

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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)