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23.1 Properties of Electric Charges 23.2 Charging Objects By Induction 23.3 Coulomb’s Law 23.4 The Electric Field 23.6 Electric Field Lines 23.7 Motion of Charged Particles in a Uniform Electric Field 1/10/2006 Norah Ali Al-moneef King Saud university 1 Electric Charge • Types: – Positive • Glass rubbed with silk • Missing electrons – Negative • Rubber/Plastic rubbed with fur • Extra electrons • Arbitrary choice – convention attributed to ? • Units: amount of charge is measured in [Coulombs] • Empirical Observations: – Like charges repel – Unlike charges attract 1/10/2006 Norah Ali Al-moneef King Saud university 2 Charge in the Atom • • • • 1/10/2006 Protons (+) Electrons (-) Ions Polar Molecules Norah Ali Al-moneef King Saud university 3 23.1 Properties of Electric Charges • Conservation electricity is the implication that electric charge is always conserved. • That is, when one object is rubbed against another, charge is not created in the process. The electrified state is due to a transfer of charge from one object to the other. • One object gains some amount of negative charge while the other gains an equal amount of positive charge. • Quantization – The smallest unit of charge is that on an electron or proton. (e = 1.6 x 10-19 C) • It is impossible to have less charge than this • It is possible to have integer multiples of this charge 1/10/2006 Q Ne Norah Ali Al-moneef King Saud university 4 23.2 Charging Objects By Induction Conductors and Insulators • Conductor transfers charge on contact • Insulator does not transfer charge on contact • Semiconductor might transfer charge on contact 1/10/2006 Norah Ali Al-moneef King Saud university 5 Charge Transfer Processes • Conduction • Polarization • Induction 1/10/2006 Norah Ali Al-moneef King Saud university 6 23-3 Coulomb’s Law • Empirical Observations F q1q 2 1 F 2 r Direction of the force is along the line joining the two charges • Formal Statement kq1q 2 F12 rˆ21 2 r21 1/10/2006 Norah Ali Al-moneef King Saud university 7 • Consider two electric charges: q1 and q2 kq1q2 • The electric force F between these two F 2 charges separated by a distance r is given by r Coulomb’s Law • The constant k is called Coulomb’s constant 9 2 2 k 910 Nm /C and is given by • The coulomb constant is also written as k 1 4 0 where 0 8.85 10 12 C2 Nm 2 • 0 is the “electric permittivity of vacuum” – A fundamental constant of nature F 1/10/2006 Norah Ali Al-moneef King Saud university 1 40 q1 q 2 r2 8 • Double one of the charges – force doubles • Change sign of one of the charges – force changes direction • Change sign of both charges – force stays the same • Double the distance between charges – force four times weaker • Double both charges – force four times stronger 1/10/2006 Norah Ali Al-moneef King Saud university 9 Example: What is the force between two charges of 1 C separated by 1 meter? Answer: 8.99 x 109 N, 1/10/2006 Norah Ali Al-moneef King Saud university 10 Coulomb’s Law Example • What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x 10-12 m • The magnitude of the Coulomb force is • F = kQ1Q2/r2 • = (9.0 x 109 N · m2/C2)(26)(1.60 x 10–19 C)(1.60 x 10–19 C)/(1.5x10–12 m)2 • = 2.7 x 10–3 N. 1/10/2006 Norah Ali Al-moneef King Saud university 11 Example - The Helium Nucleus Part 1: The nucleus of a helium atom has two protons and two neutrons. What is the magnitude of the electric force between the two protons in the helium nucleus? Answer: 58 N Part 2: What if the distance is doubled; how will the force change? Answer: 14.5 N Inverse square law: If the distance is doubled then the force is reduced by a factor of 4. 1/10/2006 Norah Ali Al-moneef King Saud university 12 Example - Equilibrium Position • Consider two charges located on the x axis x1 x2 • The charges are described by – q1 = 0.15 C – q2 = 0.35 C x = 0.0 m x = 0.40 m • Where do we need to put a third charge for that charge to be at an equilibrium point? – At the equilibrium point, the forces from the two charges will cancel. 1/10/2006 Norah Ali Al-moneef King Saud university 13 Example - Equilibrium Position (2) x1 x3 x2 • The equilibrium point must be along the x-axis. • Three regions along the x-axis where we might place our third charge x3 < x 1 x1 < x 3 < x 2 x3 > x 2 1/10/2006 Norah Ali Al-moneef King Saud university 14 Example - Equilibrium Position (3) x1 x2 • x3<x1 – Here the forces from q1 and q2 will always point in the same direction (to the left for a positive test charge) • No equilibrium • x2<x3 – Here the forces from q1 and q2 will always point in the same direction (to the right for a positive test charge) • No equilibrium 1/10/2006 Norah Ali Al-moneef King Saud university 15 Example - Equilibrium Position (4) 0.4 x x q3 x 1 < x3 < x 2 x balance. Here the forces from q1 and q2 can q1q3 q2 q3 k k 2 2 ( x) (10 x ) Answer: x3 = 0.16 m 1/10/2006 Norah Ali Al-moneef King Saud university 16 Zero Resultant Force, Example Two fixed charges, 1mC and -3mC are separated by 10cm as shown in the figure (a) where may a third charge be located so that no force acts on it? – The magnitudes of the individual forces will be equal – Directions will be opposite – Will result in a quadratic – Choose the root that gives the forces in opposite directions q1q3 q2 q3 k k 2 ( x) (10 x ) 2 X= -13.7 cm 1/10/2006 Norah Ali Al-moneef King Saud university 17 two charges are located on the positive x-axis of a coordinate system, as shown in the figure. Charge q1=2nC is 2cm from the origin, and charge q2=-3nC is 4cm from the origin. What is the total force exerted by these two charges on a charge q3=5nC located at the origin? The total force on q3 is the vector sum of the forces due to q1 and q2 individually. The total force is directed to the left, with magnitude 1.41x10-4N. 1/10/2006 Norah Ali Al-moneef King Saud university 18 Example - Charged Pendulums • Consider two identical charged balls hanging from the ceiling by strings of equal length 1.5 m (in equilibrium). Each ball has a charge of 25 C. The balls hang at an angle = 25 with y respect to the vertical. What is the mass of the balls? x Step 1: Three forces act on each ball: Coulomb force, gravity and the tension of the Ball on left : string. kq2 Fx T sin 2 d Fy T cos mg 1/10/2006 Norah Ali Al-moneef King Saud university 19 Example - Charged Pendulums (2) Step 2: The balls are in equilibrium positions. That means the sum of all forces acting on the ball is zero! T sin kq 2 / d 2 T cos mg kq mg 2 d tan 2 d=2 l sin Answer: m = 0.76 kg A similar analysis applies to the ball on the right. 1/10/2006 Norah Ali Al-moneef King Saud university 20 Electric Force and Gravitational Force • Coulomb’s Law that describes the electric force and Newton’s gravitational law have a similar functional form Felectric q1q2 k 2 r Fgravity m1m2 G 2 r • Both forces vary as the inverse square of the distance between the objects. • Gravitation is always attractive. • k and G give the strength of the force. 1/10/2006 Norah Ali Al-moneef King Saud university 21 The Hydrogen Atom Compare the electrostatic and gravitational the forces (The average separation of the electron and proton r = 5.3 x 10-11 m r F +Ze 2 19 e2 C 9 N.m 1.60 10 Fe . 2 9 10 2 2 11 40 r C 5 . 3 10 m 1 e v 2 8.2 108 N Fg G me m p r2 2 31 kg 1.67 1027 kg 11 N.m 9.1110 6.7 10 2 2 11 k g 5.3 10 m 3.6 1047 N Fe/Fg = 2 x 1039 The force of gravity is much weaker than the electrostatic force 1/10/2006 Norah Ali Al-moneef King Saud university 22 Coulomb’s Law Example y + F1 L F3 + Q Q L Q F2 x • Q = 6.0 mC • L = 0.10 m • What is the magnitude and direction of the net force on one of the charges? Q + + We find the magnitudes of the individual forces on the charge at the upper right corner: F1= F2 = kQQ/L2 = kQ2/L2 = (9 x109 N · m2/C2)(6 x10–3 C)2/(0.100 m)2 = 3.24 x107 N. F3= kQQ/(L√2)2 = kQ2/2L2 = (9 x109 N · m2/C2)(6 x10–3 C)2 /2(0.100 m)2 = 1.62 x107 N. 1/10/2006 Norah Ali Al-moneef King Saud university 23 The directions of the forces are determined from the signs of the charges and are indicated on the diagram. For the forces on the upper-right charge, we see that the net force will be along the diagonal. For the net force, we have F = F1 cos 45° + F2 cos 45° + F3 = 2(3.24x107 N) cos 45° + 1.62x107 N = 6.20 x107 N along the diagonal, or away from the center of the square. From the symmetry, each of the other forces will have the same magnitude and a direction away from the center: The net force on each charge is 6.20x107 N away from the center of the square. 1/10/2006 Norah Ali Al-moneef King Saud university 24 Example - Four Charges Consider four charges placed at the corners of a square with sides of length 1.25 m as shown on the right. What is the magnitude of the electric force on q4 resulting from the electric force from the remaining three charges? Set up an xy-coordinate system with its origin at q2. 1/10/2006 Norah Ali Al-moneef King Saud university 25 Example - Four Charges (2) Answer: F (on q4) = 0.0916 N … and the direction? 1/10/2006 Norah Ali Al-moneef King Saud university 26