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Review: Properties of Electric Charges There are two kinds of charges in nature: positive and negative; Charges of the same sign repel each other; and those of opposite sign attract one another; Total charge in an isolated system is conserved (positive = negative); Charge is quantized. Review Questions 1. 2. 3. How many types of charges in nature? A glass rod rubbed with silk will attract or repel with a rubber rubbed with fur? After a glass rod is rubbed with silk, it have a total charge of 1000 protons. How much charge does the silk have? New Concept Conductors Electrical conductors are materials in which some of the electrons are free electrons Free electrons are not bound to the atoms and can move relatively freely Examples : copper, aluminum and silver When a good conductor is charged in a small region, the charge readily distributes itself over the entire surface of the material New Concept Insulators Electrical insulators are materials in which all of the electrons are bound to atoms These electrons can not move relatively freely through the material Examples of good insulators include glass, rubber and wood When a good insulator is charged in a small region, the charge is unable to move to other regions of the material New Concept Semiconductors The electrical properties of semiconductors are somewhere between those of insulators and conductors Examples of semiconductor materials include silicon and germanium New Physical Process Charging by Induction Charging by Induction Charging by induction requires no contact with the object inducing the charge The electrons in the neutral sphere are redistributed as the negatively charged rod is approaching Charging by Induction, 2 Some electrons will leave the sphere through the ground wire There will now be more positive charges after the ground wire is removed The positive charge has been induced in the sphere “Ground” and “Negative Polarity” Pos Neg Charging by Induction, 5 The electrons remaining on the sphere redistribute themselves There is still a net positive charge on the sphere after rod is removed Charge Rearrangement in Insulators In neutral molecules, the center of positive charge coincides with that of negative charge The charges within the molecules of the material are rearranged Æa layer of charge facing the charged object Questions: What is insulator and is it useful? What is conductor and is it useful? What is semi-conductor and is it useful? Why a rubber rod rubbed with fur can absorb bits of paper? Point Charge The term point charge refers to a particle of zero size that carries an electric charge The electrical behavior of electrons and protons is well described by modeling them as point charges Proton, electron, and atomic mass Proton: proton is a particle smaller than an atom (called subatomic particle) with an electric charge of one positive fundamental unit (+1.602 × 10−19 C) and a mass of 1.6726 × 10−27 kg Electron: electron is a fundamental particle smaller than an atom (called subatomic particle) which carries a negative fundamental unit (-1.602 × 10−19 C) and a mass of 9.11x10-31 kg. 18 electrons or protons 1 C needs 6.24 x 10 -6 Typical charges can be in the µC range: 1 µC =10 C Unit of atomic mass: -27 kg 1u = 1.66x10 Problem solving Example 1: The plasma is an ionized gas. The plasma in a hot star contains quadruply ionized nitrogen atoms, N+4. Find the charge and mass of a quadruply ionized nitrogen atoms. Problem solving Example 1: The plasma is an ionized gas. The plasma in a hot star contains quadruply ionized nitrogen atoms, N+4. Find the charge and mass of a quadruply ionized nitrogen atoms. Solution: 1) Charge = 4 electrons x 1.602x10-19C =6.408x10-19C 2) Mass: mN = mN − 4 × me +4 = 14.007u − 4me = 14.007 × 1.66 × 10 27 kg − 4 × 9.11×10 −31 kg = 2.3248 ×10 − 26 kg Problem solving Example 2: Calculate the number of electrons in 1g of copper. Problem solving Example 2: Calculate the number of electrons in 1g of copper. Solution: One atom of copper has 29 electron. One atom of copper weighs 63.546u, i.e. its molar mass is 63.546 g/mol. Within 1 g of copper, the number of electrons is thus ⎛ ⎞⎛ 1.0 g atoms ⎞⎛ electrons ⎞ ⎟⎟⎜ 6.02 ×10 23 n = ⎜⎜ ⎟⎜ 29 mol ⎠⎝ ⎝ 63.546 g / mol ⎠⎝ = 2.747 ×10 23 atom ⎟ ⎠ Coulomb’s Law Charles Coulomb measured the electric forces between two small charged spheres He found that: The force is inversely proportional to the square of the separation r between the particles and directed along the line joining them. Æ Fe ∝ 1/r2. The force is proportional to the product of the charges, q1 and q2, on the two particles. Æ Fe ∝ q1q2. Fe ∝ (q1q2)/r2 The force is a conservative force (work done just depends on positions not on path). Torsion balance Coulomb’s Law, Equation Mathematically, q1 q2 Fe = ke 2 r The SI unit of charge is the coulomb (C) ke is called the Coulomb constant 9 . 2 2 ke = 8.9875 x 10 N m /C = 1/(4πεo) εo is the permittivity of free space -12 C2 / N.m2 εo = 8.8542 x 10 Vector Nature of Electric Forces In vector form, q1q2 F12 = ke 2 rˆ r r̂ is a unit vector directed from q1 to q2 The like charges produce a repulsive force between them Vector Nature of Electrical Forces, 2 Electrical forces obey Newton’s Third Law The force on q1 is equal in magnitude and opposite in direction to the force on q2 F21 = -F12 With like signs for the charges, the product q1q2 is positive and the force is repulsive Active Figure 23.7 (SLIDESHOW MODE ONLY) A Final Note about Directions The sign of the product of q1q2 gives the relative direction of the force between q1 and q2 The absolute direction is determined by the actual location of the charges The Superposition Principle The resultant force on any one charge equals the vector sum of the forces exerted by the other individual charges that are present Remember to add the forces as vectors The resultant force on q3 is the vector sum of all the forces exerted on it by other charges: F3 = F13 + F23 Problem solving Example 4: Three point charges located at the corners of a right triangle as shown. q1 = q3 = 10 μC, q2= -5 μC. a = 0.1 m. Find the total force F3 on q3 by the other two charges. F3 Problem solving Solution: The force directions and the distance between charges are shown in the figure. q q (10 ×10 C )(10 ×10 C ) = 44 N F =k = (8.99 × 10 Nm / C ) 1 13 e F23 = ke 3 9 ( 2a ) 2 −6 2 2(0.1m ) 2 q2 q3 a 2 −6 2 ( = 8.99 × 10 Nm / C 9 2 2 ) (5 ×10 C )(10 ×10 C ) = 45N −6 (0.1m )2 F3 x = F13 x + F23 x = F13 cos 45° − F23 = −13.89 N r F3 = (− 13.89 )iˆ + 31.11 ˆj r 2 2 F3 = (− 13.89 ) + (31.11) = 34.07 N −6 F3 y = F13 y + F23 y = F13 sin 45° − 0 = 31.11N Problem solving Example 4: (Think about) Case A: q1 and q2 are fixed (not movable). The force on q is zero. If pull it along x axis, what happen? If it is pulled along y axis, then what happen? Case B: If q is not negative but positive, repeat Case A. y q1 q2 + + q x Case A y + + q1 q Case B + q2 x Electric Field – Introduction Gravitational force is a field force Field forces can act through space The effect is produced even with no physical contact between objects An electric field is said to exist in the region of space around a charged object This charged object is the source charge Electric Field – Definition, cont The electric r field vector, E , at a point in space is defined r as the electric force F acting on a positive test charge, qo placed at that point divided by the test charge: r r F E= q0 Relationship Between F and E Fe = qE F = mg This is valid for a point charge only For larger objects, the field may vary over the size of the object If q is positive, F and E are in the same direction If q is negative, F and E are in opposite directions Electric Field Notes, Final The direction of E is that of the force on a positive test charge SI units: N/C Electric Field by Point Charge, Vector Form The force between the source and test charges is determined by Coulomb’s law qqo Fe = ke 2 rˆ r q Then, the electric field will be Fe q E= = ke 2 rˆ qo r r̂ is a unit vector from source to test point r̂ Electric Field Direction a) q is positive, F is directed away from q b) The direction of E is also away from the positive source charge c) q is negative, F is directed toward q d) E is also toward the negative source charge Active Figure 23.13 (SLIDESHOW MODE ONLY) Superposition with Electric Fields At any point P, the total electric field due to a group of source charges equals the vector sum of electric fields of all the charges qi E = ke ∑ 2 rˆi i ri q1 + + q3 q4 Problem solving Example 5. Three point charges are arranged as shown. (a)Find the vector electric field that the 6.00-nC and –3.00-nC charges together create at the origin. (b)Find the vector force on the 5.00-nC charge. Problem solving Example 5. Three point charges are arranged as shown. (a)Find the vector electric field that the 6.00nC and –3.00-nC charges together create at the origin. (b)Find the vector force on the 5.00-nC charge. Solution: ( ) ( )( )( ) ( )( ) ( ) 9 2 2 −9 r ke q1 × Nm C × C 8 . 99 10 / 3 . 00 10 ˆj = − 2.70 × 103 N / C ˆj E1 = 2 − ˆj = − 2 r1 ( 0.100m ) −9 9 2 2 r k e q2 8 . 99 10 / 6 . 00 10 × Nm C × C ˆ = − 5.99 × 10 2 N / C iˆ E2 = 2 − iˆ = − i 2 r ( ) 0 . 300 m 2 r r r E = E1 + E2 = − 5.99 × 10 2 N / C iˆ − 2.70 ×103 N / C ˆj r r F = qE = 5.00 ×10 −9 C − 599iˆ − 2700 ˆj N / C = − 3.00 ×10 −6 iˆ − 13.50 × 10 −6 ˆj N = − 3.00iˆ − 13.5 ˆj μN ( ) ( ( ( )( ( ) )( ) ( ) ( ) ) )