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Electric Charge and the Electric Field (Ch 26) Four Fundamental Forces 1. Electromagnetic 2. Gravity 3. Strong Nuclear Force 4. Weak Nuclear Force Electric Charge • Electricity – Elektron (amber) • Amber (tree resin) gets a charge when rubbed with cloth • Ben Franklin – Positive = Charge on a rubber rod – Negative = Charge on an amber/plastic rod Electric Charge • Law of conservation of charge – Charge is never created or destroyed Cloth - + +++ + Rubber rod Insulators/Conductors Insulators Do not conduct electricity •Small atoms •Electrons tightly held •Non-Metals Semiconductors Conduct electricity, but only at a higher voltage •Medium atoms •Electrons held medium •Metalloids Conductors Conduct electricity well •Large atoms •Electrons not tightly held •Metals (Discuss movement of the electrons) Conduction • Transfer of charge by touching • Electrons can flow between substances • Can produce a net charge on the substance Induction • Transfer of charge without touching • Electrons migrate within the substance to cause a separation of charge • Net charge on substance is still zero Induction: Grounding • Only way to produce a net charge by induction • Earth can easily absorb or donate electrons • Electrons can leave a substance, then break the ground Electroscope Charging the electroscope by induction by conduction (charge separation) (net charge) Using the electroscope • Can be used to detect the charge on a substance • (Must charge electroscope first) Coulomb’s Law • Coloumb experimented to determine magnitude of electromagnetic force • Measured the forces between charged spheres (angle of deflection). F = k Q1Q2 r2 F = Force k = 9.0 X 109 N-m/C2 (proportionality constant) Q = Charge (C) r = Radius (m) Varies with inverse-square of the radius(distance) The Coulomb • 1 Coulomb = 1 Ampere•second • Unit of charge • Point charges – small objects (charge doesn’t get distributed much) • Elementary Charge – Charge on the electron and proton – Quantized (can’t have ½ an electron) – e = 1.602 X 10-19 C Determine the magnitude of the force between a proton and an electron in a hydrogen atom. Assume the distance from the electron to the nucleus is 0.53 X 10-10 m. (8.2 X 10-8 N) + - Calculate the force between an electron and the three protons in a Li atom if the distance is about 1.3 X 10-10 m. (3.9 X 10-8 N) Three charged particles are arranged in a straight line as shown in the diagram. Calculate the net force on particle 3. (-1.5 to the left) 0.30 m Q1= -8.0 mC 0.20 m + Q2= +3.0 mC Q3 = -4.0 mC F = k Q1Q2 r2 F31 = (9.0 X 109 N-m2/C2 )(4.0X 10-6 C)(8.0X10-6 C) (0.50 m)2 F31 = 1.2 N (Repulsive to the right) F32 = (9.0 X 109 N-m2/C2 )(4.0X 10-6 C)(3.0X10-6 C) (0.20 m)2 F32 = 2.7 N (Attractive to the left) Fnet = F31 - F32 Fnet = 1.2 N – 2.7 N = -1.5 to the left Three charges are in a line. The first is +2.00 mC. The second is -2.50 mC at 25 cm. The third charge is +2.00 mC at the 40 cm mark. a) Calculate the net force on the center charge. b) Will the center charge move to the left or right? Three charges are in a line. The first is +3.00 mC. The second is -2.00 mC at 5 cm. Where should a the third charge (+4.0 mC) be placed so the middle charge does not move? Three charges are in a line. The first is -10.00 mC. The second is -15.00 mC at 100 cm. Where should a middle charge (+20.0 mC) be placed so it does not move? Coulomb’s Law: Ex 4 What is the resultant force on charge q3 if the charges are arranged as shown below. The magnitudes of the charges are: q1 q2 q3 = = = +6.00 X 10-9 C -2.00 X 10-9 C +5.00 X 10-9 C q2 - + 37o 3.00 m q1 4.00 m + 5.00 m q3 First calculate the forces on q3 separately: F13= k Q1Q3 r2 F13 = (9.0 X 109 N-m2/C2)(6.00 X 10-9 C)(5.00 X 10-9 C) (5.00m )2 F13 = 1.08 X 10-8 N F23 = k Q2Q3 r2 F23 = (9.0 X 109 N-m2/C2)(2.00 X 10-9 C)(5.00 X 10-9 C) (4.00m )2 F23 = 5.62 X 10-9 N F13 = 1.08 X 10-8 N F23 = 5.62 X 10-9 N q2 - q1 + + 37o 37o q3 F13x = F13cos37o = (1.08 X 10-8 N)cos37o F13x = 8.63 X 10-9 N F13y = F13sin37o = (1.08 X 10-8 N)sin37o F13y = 6.50 X 10-9 N Fx = F23 + F13x Fx = -5.62 X 10-9 N + 8.63 X 10-9 N = 3.01 X 10-9 N Fy = F13y = 6.50 X 10-9 N FR Fx = 3.01 X 10-9 N Fy = 6.50 X 10-9 N Fy - q FR = \/ (3.01 X 10-9 N)2 + (6.50 X 10-9 N)2 FR = 7.16 X 10-9 N sin q = Fy/FR sin q = (6.50 X 10-9 N)/ (7.16 X 10-9 N) q = 64.7o Fx Coulomb’s Law: Ex 5 Calculate the net electrostatic force on charge Q3 as shown in the diagram: + Q3 = +65 mC 60 cm 30 cm + Q2 = +50 mC 30o - Q1 = -86 mC First calculate the forces on Q3 separately: F13= k Q1Q3 r2 F13 = (9.0 X 109 N-m2/C2)(65 X 10-6 C)(86 X 10-6 C) (0.60 m )2 F13 = 140 N F23 = (9.0 X 109 N-m2/C2)(65 X 10-6 C)(50 X 10-6 C) (0.30 m )2 F23 = 330 N F23 Q3 + + Q2 F13 30o Q1 - F13x = F13cos30o = (140 N)cos30o F13x = 120 N F13y = -F13sin30o = (140 N)sin30o F13y = -70 N Fx = F13x = 120 N F7 = 330 N - 70 N = 260 N FR Fx = 120 N Fy = 260 N FR = \/ (120 N)2 + (330 N)2 FR = 290 N sin q = Fy/FR sin q = (260 N)/ (290 N) q = 64o Fy - q Fx A small plastic bead has a mass of 15 mg and a charge of -10 nC (nano = 10-9). A glass rod of charge + 10 nC is held 1.0 cm above the bead. a. Calculate the electric field strength of the rod at the position of the bead. b. Calculate the force on the bead. c. Will the bead leap off the table? Calculus Example 1 Calculate the force on charge q from the charged rod shown below. The charge per unit length of the rod is: l = Q/l F = kqQ x2 dF = kqdQ x2 dF = kqldx x2 F = kql ∫ dx x2 F = -kql 1 a+ l x a dQ = ldx (from “a” to “a+l”) F = -kql 1 x a+ l a F = -kql 1 - l a+1 a F = kqll a(a+1) Calculus Ex 2 A total positive charge of Q is evenly distributed on a semicircular ring of radius R. Calculate the force felt by the charge q at the center of the semicircle. F = kqQ R2 However, x-components cancel Fy = Fsinq Fy = kqQ sinq R2 Break into a small unit of force dFy = kqdQ sinq R2 l = Q/pR dQ = lds s = Rq dQ = l Rdq dFy = kq lR sinq dq R2 Fy = kq l - cosq R Fy = 2kq l R p 0 Electric Field • Contact forces – Friction – Pushes and pulls • Forces at a distance – Gravity – Electromagnetism • Field – Invisible lines that extend from a body Proton + + A positive test charge would be repelled by the field Electron + - A positive test charge would be attracted by the field • Opposite charges attract • Like Charges repel E in terms of a test charge E= F q • Vector quantity • Force that the test charge q would feel. The smaller the charge, the larger the Force E in terms of a point charge E = kQ r2 • Vector quantity • Point charge is the source charge producing the electric field k = 1/ 4pe0 e0 =8.85 X 10-12 C2/Nm2 (permittivity constant) Electric Field: Example 1 Find the electric force on a proton placed in an electric field of 2.0 X 104 N/C E= F q F = qE F = (1.602 X 10-19 C)(2.0 X 104 N/C) F = 3.2 X 10-15N Electric Field: Example 2 Calculate the magnitude and direction of an electric field at a point 30 cm from a source charge of Q = -3.0 X 10-6 C. E = kQ r2 E = (9.0 X 109 N-m2/C2)(3.0 X 10-6 C) (0.30 m)2 E = 3.05 X 105 N/C towards the charge What electric field is required for a copier to carry toner particles of mass 9.0 X 10-16 kg. Each particle carries 20 electrons to provide the test charge. Assume the copier must overcome twice the weight of each particle. E= F q F = 2mg (twice the weight) E = 2mg q = (2)(9.0 X 10-16 kg)(9.8m/s2) (20)(1.602X10-19C) E = 5500 N/C Electric Field: Example 3 Two point charges are separated by a distance of 10.0 cm. What is the magnitude and direction of the electric field at point P, 2.0 cm from the negative charge? 2 cm Q1 = -25 mC 8 cm P + Q2 = +50 mC E2 E1 2 cm Q1 = -25 mC 8 cm P + Q2 = +50 mC E = E1 + E2 (both point to the left) E = kQ r2 E1 = (9.0 X 109 N-m2/C2)(25 X 10-6 C) (0.020 m)2 E1 = 5.625 X 108 N/C E2 = (9.0 X 109 N-m2/C2)(50 X 10-6 C) (0.080 m)2 E2 = 7.031 X 107 N/C E = E1 + E2 = 6.3 X 108 N/C Electric Field: Example 3a What acceleration would an electron feel if it were placed at point P? Would it move to the right or the left? An electron has a mass of 9.1 X 10-31 kg. E2 E1 2 cm Q1 = -25 mC 8 cm P + Q2 = +50 mC E= F q F = ma E = ma q a = Eq/m a = (6.3 X 108 N/C)(1.602 X 10-19 C) (9.1 X 10-31 kg) a = 1.1 X 1020 m/s2 Electric Field: Example 4 Charge Q1 = 7.00 mC is placed at the origin. Charge Q2 = -5.00 mC is placed 0.300 m to the right. Calculate the electric field at point P, 0.400 m above the origin. P 0.400 m + 0.300 m Q1 = 7.00 mC Q2 = -5.00 mC P 0.400 m c = 0.500 m + 0.300 m Q1 = 7.00 mC q Q2 = -5.00 mC c2 = a2 + b2 c2 = (0.400 m)2 + (0.300 m)2 c = 0.500 m tan q = opp/adj = 0.400/0.300 q = 53.1o - E = kQ r2 E1 = (9.0 X 109 N-m2/C2)(7.00 X 10-6 C) (0.400 m)2 E1 = 3.94 X 105 N/C E2 = (9.0 X 109 N-m2/C2)(5.00 X 10-6 C) (0.500 m)2 E2 = 1.80 X 105 N/C E1 P + E2x E2 E2y q - E2x = E2cosq = (1.80 X 105 N/C)(cos 53.1o) E2x = 1.08 X 105 N/C (to the right) E2y = E2sinq = (1.80 X 105 N/C)(sin 53.1o) E2y = -1.44 X 105 N/C (down) Ex = E2x Ex = 1.08 X 105 N/C (to the right) P f Ey = E1 + E2y Ey = 3.94 X 105 N/C + -1.44 X 105 N/C Ey = 2.49 X 105 N/C ER2 = (1.08 X 105 N/C )2 + (2.49 X 105 N/C)2 ER = 2.72 X 105 N/C tan f = Ey/Ex = 2.49 X 105/ 1.08 X 105 f = 66.6o Electric Field: Example 5 Calculate the electric field at point A, as shown in the diagram A 30 cm + 52 cm Q1 = +50.0 mC Q2 = -50.0 mC Ans: E = 4.5 X 106 N/C at an angle of 76o Electric Field: Example 6 Calculate the electric field at point B, as shown in the diagram. B 30 cm + 26 cm Q1 = +50.0 mC 26 cm - Q2 = -50.0 mC Ans: E = 3.6 X 106 N/C along the +x direction