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15/05/31 Chapter 16 Electric Forces , Fields And Potentials Electric forces: Coulomb’s Law The electric force: •is inversely proportional to the square of the separation r between the particles and directed along the line joining them; • is proportional to the product of the charges q1 and q2 on the two particles; • is attractive if the charges are of opposite sign and repulsive if the charges have the same sign; • is a conservative force. 1 15/05/31 Electric forces: Coulomb’s law The Coulomb constant ke in SI units has the value Ke = 9x109 N.m2/C2 This constant is also written in the form 2 Coulomb constant 15/05/31 Example: Object A has a charge of +2 µC, and object B has a charge of +6 µC. Which statement is true about the electric forces on the objects? (a) FAB = -3FBA (b) FAB = -FBA (c) 3FAB = -FBA (d) FAB = 3FBA (e) FAB = FBA (f) 3FAB = FBA (b) FAB = -FBA Example: The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately 5.3 x 1011 m. Find the magnitudes of the electric force and the gravitational force between the two particles. (1.6 x10 19 ) 2 Fe 9 x10 x 8.2 x10 8 N 11 2 (5.3x10 ) 9 3 15/05/31 The gravitational force between charged atomic particles is negligible when compared with the electric force. Example: Consider three point charges located at the corners of a right triangle as shown in Figure, where q1 = q3 = 5.0 µC, q2 = -2.0 µC, and a = 0.10 m. Find the resultant force exerted on q3. 4 15/05/31 example: Three point charges lie along the x axis as shown in Figure. The positive charge q1 =15.0 µC is at x = 2.00 m, the positive charge q2 =6.00 µC is at the origin, and the resultant force acting on q3 is zero. What is the x coordinate of q3? 5 15/05/31 6 15/05/31 The Electric Field the electric field vector E at a point in space is defined as the electric force Fe acting on a positive test charge q0 placed at that point divided by the test charge: The Electric Field At any point P, the total electric field due to a group of source charges equals the vector sum of the electric fields of all the charges. 7 15/05/31 Example: A test charge of +3 µC is at a point P where an external electric field is directed to the right and has a magnitude of 4 x 106 N/C. If the test charge is replaced with another test charge of -3 µC, the external electric field at P (a) is unaffected (b) reverses direction (c) changes in a way that cannot be determined (b) reverses direction Electric field diagrams •The electric field vector E is tangent to the electric field line at each point. The line has a direction, indicated by an arrowhead, that is the same as that of the electric field vector. • The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region. Thus, the field lines are close together where the electric field is strong and far apart where the field is weak. •The lines must begin on a positive charge and terminate on a negative charge. In the case of an excess of one type of charge, some lines will begin or end infinitely far away. • The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge. • No two field lines can cross 8 15/05/31 Electric field diagrams Electric field diagrams 9 15/05/31 Electric field diagrams Electric Field .B Rank the magnitudes E of the electric field at points A, B, and C shown in the figure. A) EC>EB>EA B) EB>EC>EA C) EA>EC>EB D) EB>EA>EC E) EA>EB>EC 10 .C .A 15/05/31 Example: Example: A charge q1 = 7.0 µC is located at the origin, and a second charge q2 = -5.0 µC is located on the x axis, 0.30 m from the origin. Find the electric field at the point P, which has coordinates (0, 0.40) m. 11 15/05/31 Significance of the electric field 12 15/05/31 The Electric Field Significance of the electric field 13 15/05/31 Problem #2: Calculate the electric field at point P below. Charge A = 6C Charge B = 2C Distance from A to P = .4 meters Distance from P to B = .2 meters A .4 P .2 6C B 2C EB= (9x109)(2) .22 EA= (9x109)(6) .42 E=4.5x1011 N/C E= 3.375x1011 N/C Since the arrows point opposite directions, subtract the two magnitudes to find the net electric field A positive test charge placed at point P would move to the right, away from charge A A positive test charge placed at point P would move to the left, away from charge B 3.375x1011 4.5x1011 4.5 x 1011 3.375 x 10111.125 x 1011 N/C The resulting direction would be to the left since that electric field was stronger 14 15/05/31 Electric Field of a Dipole Start with 1 q 1 q 4 0 r2 4 0 r2 1 q 1 q 4 0 ( z d / 2) 2 4 0 ( z d / 2) 2 q d d [(1 ) 2 (1 ) 2 ] 4 0 z 2 2z 2z If d << z, then, [(1 E E E So d 2 d 2d 2d 2d ) (1 ) 2 ] [(1 ...) (1 ...)] 2z 2z 2 z (1!) 2 z (1!) z E q 2d 1 qd 4 0 z 2 z 2 0 z 3 p qd 1 p E 2 0 z 3 E ~ 1/z3 E =>0 as d =>0 Valid for “far field” Spherical Conductors Because it is conducting, charge on a metal sphere will go everywhere over the surface. You can easily see why, because each of the charges pushes on the others so that they all move apart as far as they can go. Because of the symmetry of the situation, they spread themselves out uniformly. There is a theorem that applies to this case, called the shell theorem, that states that the sphere will act as if all of the charge were concentrated at the center. Note, forces are equal and opposite These two situations are the same 15 15/05/31 Electric charge Condense charge E = (kQ/r2) r̂ outside spherically symmetric charge distribution Properties of Electric fields 16 Concentric spherical shells: 4kQ 4kQ E Field just outside sphere A 4R 2 15/05/31 Uniformly charge planes E 2kQ xˆ A Uniformly charge planes E 4kQ xˆ A Example 16.3/384 17 15/05/31 The electrical potential B U U B U A qE.ds qEl A V VB V A El l Oppositely charged plates V 4kQ l A l The potential due to a point charge: 18 15/05/31 Example: A charge q1 = 2.00 µC is located at the origin, and a charge q2 = -6.00 µC is located at (0, 3.00) m, as shown in Figure. (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. Figure a (B) Find the change in potential energy of the system of two charges plus a charge q3 = 3.00 µC as the latter charge moves from infinity to point P. Figure b A) 19 15/05/31 B) Capacitance: The capacitance C of a capacitor is defined as the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between the conductors: C 20 Q V Farad (F) 15/05/31 The parallel plate capacitor: C A 4kd Example: A parallel-plate capacitor with air between the plates has an area A = 2.00 x 10-4 m2 and a plate separation d = 1.00 mm. Find its capacitance. 21