Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Transcript

Physics 6B Capacitors Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 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 C2 Voltage across C1 and C2 must be equal. Charge on each may be different. 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. Voltage across C1 and C2 may be different. Charge on each must be equal. 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. + + + + + + + + + + + + + 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. + + + + + + + + + + + + 0.75 cm + _ 12 V _ V = E∙d 12 V E (0.0075 m) E 1600 + _ _ _ _ _ _ _ _ _ _ _ _ V m 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. For a parallel-plate capacitor we have a very simple formula relating the voltage to the electric field inside. + + + + + + + + + + + + E 0.75 cm + _ 12 V _ 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 - ) 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. + + + + + + + + + + + + E 0.75 cm + _ 12 V _ 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 - ) 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. For a parallel-plate capacitor we have a very simple formula relating the voltage to the electric field inside. + + + + + + + + + + + + E 0.75 cm + _ 12 V _ 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 - ) 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 (6.24 106 C)(12V) 7.49 105 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. 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. K 7.49 105 J 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. K 7.49 105 J We can now calculate the mass from our definition of kinetic energy. K 12 m v2 7.49 105 J 12 m (3.4 ms )2 m 1.3 105kg 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. C1=6μF; C2=2μF C1 6V C2 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. C1=6μF; C2=2μF C1 These capacitors are in series. Use the formula to find the equivalent capacitance: Ceq 6F 2F 12 3 F F 6F 2F 8 2 6V C2 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. C1=6μF; C2=2μF C1 These capacitors are in series. Use the formula to find the equivalent capacitance: Ceq 6F 2F 12 3 F F 6F 2F 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 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. C1=6μF; C2=2μF C1 These capacitors are in series. Use the formula to find the equivalent capacitance: Ceq 6F 2F 12 3 F F 6F 2F 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) 9C 2 6V C2 Ceq 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. C1=6μF; C2=2μF C1 These capacitors are in series. Use the formula to find the equivalent capacitance: Ceq 6F 2F 12 3 F F 6F 2F 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) 9C 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 For Campus Learning 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 6F 2F 12 3 F F 6F 2F 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) 9C 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 9C 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. C1=6μF; C2=2μF C1 These capacitors are in series. Use the formula to find the equivalent capacitance: Ceq 6F 2F 12 3 F F 6F 2F 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) 9C 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 9C Rearranging our basic formula and applying it to each individual capacitor gives us the voltage across each: 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. C1=6μF; C2=2μF C1 These capacitors are in series. Use the formula to find the equivalent capacitance: Ceq 6V 6F 2F 12 3 F F 6F 2F 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) 9C 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 9C 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 9C 3 Volts C1 6F 2 V2 Q2 9C 9 Volts C2 2F 2 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. C1=6μF; C2=2μF C1 These capacitors are in series. Use the formula to find the equivalent capacitance: Ceq 6V 6F 2F 12 3 F F 6F 2F 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) 9C 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 9C 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 9C 3 Volts C1 6F 2 Our final calculations use a formula for stored energy: V2 Q2 9C 9 Volts C2 2F 2 3 27 Uelec,1 12 Q1V1 12 (9C)( V) J 2 4 9 81 Uelec,2 12 Q2 V2 12 (9C)( V) J 2 4 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. C1=6μF; C2=2μF C1 These capacitors are in series. Use the formula to find the equivalent capacitance: Ceq 6V 6F 2F 12 3 F F 6F 2F 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) 9C 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 9C 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 9C 3 Volts C1 6F 2 Our final calculations use a formula for stored energy: V2 Q2 9C 9 Volts C2 2F 2 3 27 Uelec,1 12 Q1V1 12 (9C)( V) J 2 4 9 81 Uelec,2 12 Q2 V2 12 (9C)( 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. 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 6V C3 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 We need to find the equivalent capacitance for this circuit, then work backwards to find the energy in each capacitor. C1 C2 6V C3 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 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 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 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 3F 3F 9 3 F F 3F 3F 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 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 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 3F 3F 9 3 F F 3F 3F 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) 9C 2 Next we will work backwards to find the information about each individual 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=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 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. 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 9C 3Volts C3 3F 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 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 9C 3Volts C3 3F We can also find the energy stored in C3, as well as the total. 27 J 2 12 (9C)(6V) 27J U3 12 Q3V3 12 (9C)(3V) Utotal 12 QtotalVtotal 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 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 (1F)(3v) 3c Q2 C2V2 (2F)(3v) 6c 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 (6C)(3V) 9J Ceq U1 12 Q1V1 12 (3C)(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 (6C)(3V) 9J Ceq U1 12 Q1V1 12 (3C)(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