15/20/2015 General Physics (PHY 2140) Lecture 6  Electrostatics and electrodynamics Capacitance and capacitors capacitors with dielectrics Electric current current and drift speed resistance and Ohm’s law Chapter

25/20/2015 Lightning Review Last lecture: 1.Capacitance and capacitors Parallel-plate capacitor Parallel-plate capacitor Combinations of capacitors Combinations of capacitors Parallel Parallel Series Series Energy stored in a capacitor Energy stored in a capacitor Review Problem: Consider an isolated simple parallel-plate 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.

35/20/2015 QQ QQ QQ QQ V0V0 V 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

45/20/2015 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 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

55/20/2015 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 For a more complete list, see Table 16.1

65/20/2015 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. Example

75/20/2015 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 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

85/20/2015 How does an insulating dielectric material reduce electric fields by producing effective surface charge densities?                                                                 Reorientation of polar molecules Induced polarization of non-polar molecules                                 Dielectric Breakdown: breaking of molecular bonds/ionization of molecules.

95/20/ 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

105/20/ 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. A I

115/20/ Electric Current - Units The SI units of current is the ampere (A). 1 A = 1 C/s 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. 1 A of current is equivalent to 1 C of charge passing through the area in a time interval of 1 s.

125/20/ 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.

135/20/ 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. The direction of the current in copper is thus opposite the direction of the electrons v I

145/20/ 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.

155/20/ 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.

165/20/ 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.

175/20/2015 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.

185/20/2015 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 cross-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.

195/20/2015 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:

205/20/ Current and Drift Speed Consider the current on a conductor of cross-sectional area A. A vdvd q vdtvdt

215/20/ Current and Drift Speed (2) Volume of an element of length  x is :  V = A  x. Let n be the number of carriers per unit of volume. The total number of carriers in  V is: n A  x. The charge in this volume is:  Q = (n A  x)q. Distance traveled at drift speed v d by carrier in time  t:  x = v d  t. Hence:  Q = (n A v d  t)q. The current through the conductor: I =  Q/  t = n A v d q.

225/20/ 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.

235/20/ Current and Drift Speed - Example Question: A copper wire of cross-sectional area 3.00x10 -6 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.

245/20/2015 Question: A copper wire of cross-sectional area 3.00x10 -6 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 = 3.00x10 -6 m 2 I = 10 A.  = 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.

255/20/2015 Question: A copper wire of cross-sectional area 3.00x10 -6 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 = 3.00x10 -6 m 2 ; I = 10 A.;  = 8.95 g/cm 3.; q = 1.6 x C. n = 8.48 x electrons/ cm 3.

265/20/ Current and Drift Speed - Comments Drift speeds are usually very small. Drift speed much smaller than the average speed between collisions. Electrons traveling at 2.46x10 -6 m/s would would take 68 min to travel 1m. Electrons traveling at 2.46x10 -6 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… The info travels at roughly 10 8 m/s…

275/20/2015 Mini-quiz 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.

285/20/ 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. VV I

295/20/ Definition of Resistance In situations where the proportionality is exact, one can write. The proportionality constant R is called resistance of the conductor. The resistance is defined as the ratio.

305/20/ Resistance - Units In SI, resistance is expressed in volts per ampere. A special name is given: ohms (  ). 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 .

315/20/ 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 ( )

325/20/2015 I VV I VV Linear or Ohmic Material Non-Linear or Non-Ohmic Material Semiconductors e.g. diodes Most metals, ceramics

335/20/2015 Ohm’s Law R understood to be independent of  V.

345/20/2015 Definition: Resistor: a conductor that provides a specified resistance in an electric circuit. The symbol for a resistor in circuit diagrams.

355/20/2015 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?