Créer une présentation
Télécharger la présentation

Télécharger la présentation
## Electric Charge and the Electric Field (Ch 26)

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Electric Charge and the Electric Field (Ch 26)**Four Fundamental Forces • Electromagnetic • Gravity • Strong Nuclear Force • 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**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**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 0.20 m - - + Q1= -8.0 mC 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. • Calculate the net force on the center charge. • 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 q3if the charges are arranged as shown below. The magnitudes of the charges are: q1 = +6.00 X 10-9 C q2 = -2.00 X 10-9 C q3 = +5.00 X 10-9 C**4.00 m**q2 - + q3 37o 3.00 m 5.00 m q1 +**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 37o q2 - + q3 37o q1 +**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**Fy Fx= 3.01 X 10-9 N Fy= 6.50 X 10-9 N 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 q 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 30o + - Q2= +50 mC 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 + F13 30o + - Q2 Q1**F13x = F13cos30o = (140 N)cos30o**F13x = 120N F13y = -F13sin30o = (140 N)sin30o F13y = -70N Fx = F13x = 120 N F7 = 330 N - 70N = 260 N**FR**Fy Fx= 120 N Fy= 260N FR = \/ (120 N)2 + (330N)2 FR = 290 N sin q = Fy/FR sin q = (260N)/ (290N) q = 64o 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. • Calculate the electric field strength of the rod at the position of the bead. • Calculate the force on the bead. • 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 = kqdQdQ = ldx x2 dF = kqldx x2 F = kql∫ dx (from “a” to “a+l”) x2 F = -kql1 a+ l x a**F = -kql1 a+ l**x a F = -kql1 - 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 p**R 0 Fy = 2kq l R**Electric Field**• Contact forces • Friction • Pushes and pulls • Forces at a distance • Gravity • Electromagnetism • Field – Invisible lines that extend from a body**A positive test charge would be repelled by the field**+ Proton +**A positive test charge would be attracted by the field**+ Electron -**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)