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: Voltage Source + _ _ _ _ _ _ _ _ _ _ _ _ _ _ d E Diagram of a parallel-plate capacitor Capacitors in Parallel: Voltage across C 1 and C 2 must be equal. Charge on each may be different. C1C1 C2C2 Capacitors in Series: Voltage across C 1 and C 2 may be different. Charge on each must be equal. C1C1 C2C2 Shortcut – works for any pair of capacitors in series.

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 x 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 x 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 x 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 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 x 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 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 x 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 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 x 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 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: 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.

Prepared by Vince Zaccone 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?

Prepared by Vince Zaccone 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.

Prepared by Vince Zaccone 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. We can now calculate the mass from our definition of kinetic energy.

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 C2C2 C1C1 6V

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: Circuit Example #1: Find the voltage across, and energy stored in each capacitor in the circuit shown. C 1 =6μF; C 2 =2μF

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: 6V C eq 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. C 1 =6μF; C 2 =2μF

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: 6V C eq 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. C 1 =6μF; C 2 =2μF

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: 6V C eq The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge: 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. C 1 =6μF; C 2 =2μF

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: 6V C eq The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge: 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. C 1 =6μF; C 2 =2μF

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: 6V C eq The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge: 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). 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. C 1 =6μF; C 2 =2μF

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: 6V C eq The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge: 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). 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. Circuit Example #1: Find the voltage across, and energy stored in each capacitor in the circuit shown. C 1 =6μF; C 2 =2μF

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: 6V C eq The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge: 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). 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. Our final calculations use a formula for stored energy: Circuit Example #1: Find the voltage across, and energy stored in each capacitor in the circuit shown. C 1 =6μF; C 2 =2μF

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C2C2 C1C1 6V These capacitors are in series. Use the formula to find the equivalent capacitance: 6V C eq The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge: 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). 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. Our final calculations use a formula for stored energy: 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. C 1 =6μF; C 2 =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. C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 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. C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 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. C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 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 C 1 and C 2 are in parallel to each other, so they are equivalent to a single capacitor with capacitance C 1 +C 2 =3µF. Draw a new diagram for this: C3C3 C 1 +C 2 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. C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 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 C 1 and C 2 are in parallel to each other, so they are equivalent to a single capacitor with capacitance C 1 +C 2 =3µF. Draw a new diagram for this: C3C3 C 1 +C 2 6V 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: 6V C eq 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. C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 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 C 1 and C 2 are in parallel to each other, so they are equivalent to a single capacitor with capacitance C 1 +C 2 =3µF. Draw a new diagram for this: C3C3 C 1 +C 2 6V 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: 6V C eq The new diagram has just a single capacitor. Now we can use the definition of capacitance to find the charge: Next we will work backwards to find the information about each individual capacitor:

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 6V C3C3 C 1 +C 2 6V C eq Capac.VoltageChargeEnergy C1C1 1µF C2C2 2µF C3C3 3µF C eq 1.5µF6V9µC 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 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. 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 C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 6V C3C3 C 1 +C 2 6V C eq Capac.VoltageChargeEnergy C1C1 1µF C2C2 2µF C3C3 3µF9µC C eq 1.5µF6V9µC 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 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: 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 C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 6V C3C3 C 1 +C 2 6V C eq Capac.VoltageChargeEnergy C1C1 1µF C2C2 2µF C3C3 3µF3V9µC C eq 1.5µF6V9µC 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 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: We can also find the energy stored in C 3, as well as the total. 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 C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 6V C3C3 C 1 +C 2 6V C eq Capac.VoltageChargeEnergy C1C1 1µF C2C2 2µF C3C3 3µF3V9µC13.5µJ C eq 1.5µF6V9µC27µJ 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 C 1 and C 2. 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.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 6V C3C3 C 1 +C 2 6V C eq Capac.VoltageChargeEnergy C1C1 1µF C2C2 2µF C3C3 3µF3V9µC13.5µJ C eq 1.5µF6V9µC27µJ 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 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 3 (in series with the others) is 3V, that leaves 3V left for C 1 and C 2. 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.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 6V C3C3 C 1 +C 2 6V C eq Capac.VoltageChargeEnergy C1C1 1µF3V C2C2 2µF3V C3C3 3µF3V9µC13.5µJ C eq 1.5µF6V9µC27µJ 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 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 3 (in series with the others) is 3V, that leaves 3V left for C 1 and C 2. 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 C 1 and C 2 as well: 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 C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 6V C3C3 C 1 +C 2 6V C eq Capac.VoltageChargeEnergy C1C1 1µF3V3µC C2C2 2µF3V6µC C3C3 3µF3V9µC13.5µJ C eq 1.5µF6V9µC27µJ 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 C 1 and C 2, and we are done. 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 C1C1 C 1 =1μF; C 2 =2μF; C 3 =3μF C2C2 C3C3 6V C3C3 C 1 +C 2 6V C eq Capac.VoltageChargeEnergy C1C1 1µF3V3µC4.5µJ C2C2 2µF3V6µC9µJ C3C3 3µF3V9µC13.5µJ C eq 1.5µF6V9µC27µJ 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 C 1 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. 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 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)