General Physics (PHY 2140)


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1 General Physics (PHY 2140) Lecture 7 Electrostatics and electrodynamics Capacitance and capacitors capacitors with dielectrics Electric current current and drift speed resistance and Ohm s law Chapter
2 Reminder (for those who don t read syllabus) Reading Quizzes (bonus 5%): It is important for you to come to class prepared, i.e. be familiar with the material to be presented. To test your preparedness, a simple fiveminute quiz, testing your qualitative familiarity with the material to be discussed in class, will be given at the beginning of some of the classes. No makeup reading quizzes will be given. There could be one today but then again 2
3 Lightning Review Last lecture: 1. Capacitance and capacitors Q C = V A Parallelplate capacitor C = ε0 d Combinations of capacitors Ceq = C1 C2... Parallel Series =... Energy stored in a capacitor Ceq C1 C2 2 1 Q 1 U = QV = = CV 2 2C 2 Review Problem: Consider an isolated simple parallelplate capacitor whose plates are given equal and opposite charges and are separated by a distance d. Suppose the plates are pulled apart until they are separated by a distance D>d. The electrostatic energy stored in a capacitor is a. greater then b. the same as c. smaller then before the plates were pulled apart. 3 2
4 16.10 Capacitors with dielectrics A dielectrics is an insulating material (rubber, glass, etc.) Consider an insolated, charged capacitor Q Q Q Q Insert a dielectric V 0 V Notice that the potential difference decreases (k = V 0 /V) Since charge stayed the same (Q=Q 0 ) capacitance increases C Q Q κq = = = = V V κ V 0 0 κc 0 dielectric constant: k = C/C 0 Dielectric constant is a material property 4
5 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 parallelplate plate capacitor A C = κε 0 d The capacitance is limited from above by the electric discharge that can occur through the dielectric material separating the plates In other words, there exists a maximum of the electric field, sometimes called dielectric strength,, that can be produced in the dielectric before it breaks down 5
6 Dielectric constants and dielectric strengths of various materials at room temperature Material Vacuum Dielectric constant, k 1.00 Dielectric strength (V/m)  Air Water Fused quartz For a more complete list, see Table
7 Example Take a parallel plate capacitor whose plates have an area of 2.0 m 2 and are separated by a distance of 1mm. 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. 7
8 Take a parallel plate capacitor whose plates have an area of 2 m 2 and are separated by a distance of 1mm. 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 parallelplate capacitor, the original capacitance can be found as 2 A m C0 = ε0 = ( C N m ) = 18 nf 3 d m The dielectric constant and the new capacitance are V C = κc = C = 318 nf = 54nF V2 The charge on the capacitor can be found to be ( )( ) 9 5 Q= C V = F V = C
9 How does an insulating dielectric material reduce electric fields by producing effective surface charge densities? Reorientation of polar molecules Induced polarization of nonpolar molecules Dielectric Breakdown: breaking of molecular bonds/ionization of molecules. 9
10 17.1 Electric Current Whenever charges of like signs move in a given direction, a current is said to exist. Consider charges are moving perpendicularly to a surface of area A. Definition: the current is the rate at which charge flows through this surface. A I 10
11 17.1 Electric Current  Definition Given an amount of charge, Q, passing through the area A in a time interval t, the current is the ratio of the charge to the time interval. I = Q t A I 11
12 17.1 Electric Current  Units The SI units of current is the ampere (A). 1 A = 1 C/s 1 A of current is equivalent to 1 C of charge passing through the area in a time interval of 1 s. 12
13 17.1 Electric Current Remark 1 Currents may be carried by the motion of positive or negative charges. It is conventional to give the current the same direction as the flow of positive charge. 13
14 17.1 Electric Current Remark 2 In a metal conductor such as copper, the current is due to the motion of the electrons (negatively charged). The direction of the current in copper is thus opposite the direction of the electrons v I 14
15 17.1 Electric Current Remark 3 In a beam of protons at a particle accelerator (such as RHIC at Brookhaven national laboratory), the current is the same direction as the motion of the protons. In gases and electrolytes (e.g. Car batteries), the current is the flow of both positive and negative charges. 15
16 17.1 Electric Current Remark 4 It is common to refer to a moving charge as a mobile charge carrier. In a metal the charge carriers are electrons. In other conditions or materials, they may be positive or negative ions. 16
17 17.1 Electric Current Example Current in a light bulb The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find. (A) the current in the light bulb. (B) the number of electrons that pass through the filament in 1 second. 17
18 The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find. (A) the current in the light bulb. I Q 1.67C = = = t 2.00s 0.835A 18
19 The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find. (b) the number of electrons that pass through the filament in 1 second. Reasoning: In 1 s, C of charge passes the crosssectional sectional area of the filament. This total charge per second is equal to the number of electrons, N, times the charge on a single electron. 19
20 The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find. (b) the number of electrons that pass through the filament in 1 second. Solution: ( 19 ) N = N C / electron = 0.835C q N N = 0.835C C/ electron = electrons 20
21 15.2 Current and Drift Speed Consider the current on a conductor of crosssectional sectional area A. A q v d v d t 21
22 15.2 Current and Drift Speed (2) Volume of an element of length x is : V V = A x. Let n be the number of carriers per unit of volume. The total number of carriers in V V is: n A x. The charge in this volume is: Q Q = (n A x)q. Distance traveled at drift speed v d by carrier in time t: x = v d t. Hence: Q Q = (n A v d t)q. The current through the conductor: I = Q/ t t = n A v d q. 22
23 15.2 Current and Drift Speed (3) In an isolated conductor, charge carriers move randomly in all directions. When an external potential is applied across the conductor, it creates an electric field inside which produces a force on the electron. Electrons however still have quite a random path. As they travel through the material, electrons collide with other electrons, and nuclei, thereby losing or gaining energy. The work done by the field exceeds the loss by collisions. The electrons then tend to drift preferentially in one direction. 23
24 15.2 Current and Drift Speed  Example Question: A copper wire of crosssectional sectional area 3.00x106 m 2 carries a current of 10. A. Assuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The e density of copper is 8.95 g/cm 3. 24
25 Question: A copper wire of crosssectional area 3.00x106 m 2 carries a current of 10 A. Assuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The density of copper is 8.95 g/cm 3. Reasoning: We know: A = 3.00x106 m 2 I = 10 A. ρ = 8.95 g/cm 3. q = 1.6 x C. n = 6.02x10 23 atom/mol x 8.95 g/cm 3 x ( 63.5 g/mol) 1 n = 8.48 x electrons/ cm 3. 25
26 Question: A copper wire of crosssectional area 3.00x106 m 2 carries a current of 10 A. Assuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The density of copper is 8.95 g/cm 3. Ingredients: A = 3.00x106 m 2 ; I = 10 A.; ρ = 8.95 g/cm 3.; q = 1.6 x C. n = 8.48 x electrons/ cm 3. v d I 10.0 C/ s = = nqa electrons m C m = / ( 22 3 )( 19 )( ) m s 26
27 15.2 Current and Drift Speed  Comments Drift speeds are usually very small. Drift speed much smaller than the average speed between collisions. Electrons traveling at 2.46x106 m/s would would take 68 min to travel 1m. So why does light turn on so quickly when one flips a switch? The info travels at roughly 10 8 m/s 27
28 Miniquiz Consider a wire has a long conical shape. How does the velocity of the electrons vary along the wire? Every portion of the wire carries the same current: as the cross sectional area decreases, the drift velocity must increase to carry the same value of current. This is dues to the electrical field lines being compressed into a smaller area, thereby increasing the strength of the electric field. 28
29 17.3 Resistance and Ohm s Law  Intro When a voltage (potential difference) is applied across the ends of a metallic conductor, the current is found to be proportional to the applied voltage. I V V I 29
30 17.3 Definition of Resistance In situations where the proportionality is exact, one can write. V = IR The proportionality constant R is called resistance of the conductor. The resistance is defined as the ratio. R = V I 30
31 17.3 Resistance  Units In SI, resistance is expressed in volts per ampere. A special name is given: ohms (Ω).( Example: if a potential difference of 10 V applied across a conductor produces a 0.2 A current, then one concludes the conductors has a resistance of 10 V/0.2 a = 50 Ω. 31
32 17.3 Ohm s Law Resistance in a conductor arises because of collisions between electrons and fixed charges within the material. In many materials, including most metals, the resistance is constant over a wide range of applied voltages. This is a statement of Ohm s law. Georg Simon Ohm ( ) 32
33 Linear or Ohmic Material I NonLinear or NonOhmic Material I V V Most metals, ceramics Semiconductors e.g. diodes 33
34 Ohm s Law V = IR R understood to be independent of V. 34
35 Definition: Resistor: a conductor that provides a specified resistance in an electric circuit. The symbol for a resistor in circuit diagrams. 35
36 Example: Resistance of a Steam Iron All household electric devices are required to have a specified resistance (as well as many other characteristics ). Consider that the plate of a certain steam iron states the iron carries a current of 7.40 A when connected to a 120 V source. What is the resistance of the steam iron? V 120V R = = = 16.2Ω I 7.40A 36
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