111/16/2015 ELECTRICITY AND MAGNETISM Phy 220 Chapter 4: Capacitors

211/16/2015 a +Q b -Q 4.1 The definition of capacitance Capacitor: two conductors (separated by an insulator) usually oppositely charged The capacitance, C, of a capacitor is defined as a ratio of the magnitude of a charge on either conductor to the magnitude of the potential difference between the conductors

311/16/ A capacitor is basically two parallel conducting plates with insulating material in between. The capacitor doesn’t have to look like metal plates. 2.When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates. 3.Capacitors in circuits symbols: 4.Analysis follow from conservation of energy and conservation of charge Capacitor for use in high-performance audio systems

411/16/2015 Units of capacitance The unit of C is the farad (F), but most capacitors have values of C ranging from picofarads to microfarads (pF to  F). Recall, micro  10 -6, nano  10 -9, pico  If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy).

511/16/2015 A A +Q -Q d 4.2 The parallel-plate capacitor The capacitance of a device depends on the geometric arrangement of the conductors where A is the area of one of the plates, d is the separation,  0 is a constant called the permittivity of free space,  0 = 8.85  C 2 /N·m 2  0 = 8.85  C 2 /N·m 2

611/16/2015 Proof: According to example 2.4.3,chapter 2, the value of the electric field between the plates is Because the field between the plates is uniform, the magnitude of the potential difference between the plates equals Ed, Because the field between the plates is uniform, the magnitude of the potential difference between the plates equals Ed, therefore, So we find that the capacitance is

711/16/2015 A parallel plate capacitor has plates 2.00 m 2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate Problem: parallel-plate capacitor

811/16/2015 A parallel plate capacitor has plates 2.00 m 2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate Given:  V=10,000 V A = 2.00 m 2 d = 5.00 mm Find: C=? Q=? Solution: Since we are dealing with the parallel-plate capacitor, the capacitance can be found as Once the capacitance is known, the charge can be found from the definition of a capacitance via charge and potential difference:

911/16/ Combinations of capacitors It is very often that more than one capacitor is used in an electric circuit We would have to learn how to compute the equivalent capacitance of certain combinations of capacitors C1C1 C2C2 C3C3 C5C5 C1C1 C2C2 C3C3 C4C4

1011/16/2015 +Q 1 Q1Q1 C1C1 V=V ab a b +Q 2 Q2Q2 C2C2 a. Parallel combination Connecting a battery to the parallel combination of capacitors is equivalent to introducing the same potential difference for both capacitors, A total charge transferred to the system from the battery is the sum of charges of the two capacitors, By definition, Thus, C eq would be

1111/16/2015 Parallel combination: notes Analogous formula is true for any number of capacitors, It follows that the equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors (parallel combination)

1211/16/2015 A 3  F capacitor and a 6  F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. Problem: parallel combination of capacitors

1311/16/2015 A 3  F capacitor and a 6  F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. +Q 1 Q1Q1 C1C1 V=V ab a b +Q 2 Q2Q2 C2C2 Given: V = 18 V C 1 = 3  F C 2 = 6  F Find: C eq =? Q=? First determine equivalent capacitance of C 1 and C 2 : Next, determine the charge

1411/16/2015 b. Series combination Connecting a battery to the serial combination of capacitors is equivalent to introducing the same charge for both capacitors, A voltage induced in the system from the battery is the sum of potential differences across the individual capacitors, By definition, Thus, C eq would be +Q 1 Q1Q1 C1C1 +Q 2 Q2Q2 C2C2 V=V ab a c b

1511/16/2015 Series combination: notes Analogous formula is true for any number of capacitors, It follows that the equivalent capacitance of a series combination of capacitors is always less than any of the individual capacitance in the combination (series combination)

1611/16/2015 A 3  F capacitor and a 6  F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance. Problem: series combination of capacitors

1711/16/2015 A 3  F capacitor and a 6  F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited. +Q 1 Q1Q1 C1C1 +Q 2 Q2Q2 C2C2 V=V ab a c b Given: V = 18 V C 1 = 3  F C 2 = 6  F Find: C eq =? Q=? First determine equivalent capacitance of C 1 and C 2 : Next, determine the charge

1811/16/ Energy stored in a charged capacitor The energy (U) stored on a capacitor is given by:

1911/16/2015   Find electric field energy density (energy per unit volume) in a parallel-plate capacitor Example: electric field energy in parallel-plate capacitor Recall Thus, and so, the energy density is

2011/16/2015 C1C1 C2C2 V C3C3 In the circuit shown V = 48V, C 1 = 9  F, C 2 = 4  F and C 3 = 8  F. (a)determine the equivalent capacitance of the circuit, (b) determine the energy stored in the combination by calculating the energy stored in the equivalent capacitance. Example: stored energy

2111/16/2015 C1C1 C2C2 V C3C3 In the circuit shown V = 48V, C 1 = 9  F, C 2 = 4  F and C 3 = 8  F. (a) determine the equivalent capacitance of the circuit, (b) determine the energy stored in the combination by calculating the energy stored in the equivalent capacitance, First determine equivalent capacitance of C 2 and C 3 : The energy stored in the capacitor C 123 is then Next, determine equivalent capacitance of the circuit by noting that C 1 and C 23 are connected in series Given: V = 48 V C 1 = 9  F C 2 = 4  F C 3 = 8  F Find: C eq =? U=?

2211/16/2015 QQ QQ QQ QQ V0V0 V 4.5 Capacitors with dielectrics A dielectrics is an insulating material (rubber, glass, etc.) Consider an insolated, charged capacitor Notice that the potential difference decreases (k = V 0 /V) Since charge stayed the same (Q=Q 0 ) → capacitance increases dielectric constant: k = C/C 0 Dielectric constant is a material property Insert a dielectric

2311/16/2015 Capacitors with dielectrics - notes Capacitance is multiplied by a factor k when the dielectric fills the region between the plates completely E.g., for a parallel-plate capacitor There exists a maximum of the electric field, sometimes called dielectric strength, that can be produced in the dielectric before it breaks down

2411/16/2015 Dielectric constants and dielectric strengths of various materials at room temperature Material Dielectric constant, k Dielectric strength (V/m) Vacuum Air  10 6 Water80-- Fused quartz  10 6

2511/16/2015 Take a parallel plate capacitor whose plates have an area of 2000 cm 2 and are separated by a distance of 1cm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kV. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material. Example

2611/16/2015 Take a parallel plate capacitor whose plates have an area of 2 m 2 and are separated by a distance of 1cm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kV. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material. Given:  V 1 =3,000 V  V 2 =1,000 V A = 2.00 m 2 d = 0.01 m Find: C=? C 0 =? Q=? k=? Since we are dealing with the parallel-plate capacitor, the original capacitance can be found as The charge on the capacitor can be found to be The dielectric constant and the new capacitance are