Unit 7 Electrostatics Lesson 1 Electric Charge Electric

Unit 7: Electrostatics Lesson 1: Electric Charge

Electric Charge Smallest bits of electric charge: protons and electrons e = elementary charge (charge on one proton/electron) = 1. 6 x 10 -19 Coulombs E. g. How many electrons in one coulomb? 1 C / e = 6. 2 x 1018 electrons

Laws of Electric Charge Like charges repel; opposite charges attract. Charged objects (negative or positive) attract neutral ones. E. g. A negative object repels the electrons at the surface of a neutral object brought near it, thus creating a positive surface that attracts to it. This is called “charging by induction”. Negative object Neutral object

Movement of Electrons Charging by contact: transfer of charge when electrons move from place to place Charging by induction: temporary movement of electrons within an object due to electric forces Demo:

Brain Break!

Like charges repel; unlike charges attract… with how much force? q 1 d q 2 Point charges q = magnitude of charge in coulombs Coulomb’s Law: F = k q 1 q 2 d 2 k = 9. 0 x 109 Nm 2/C 2 “Coulomb’s constant”

Example 1: An electron orbits a proton at a distance of 2. 0 x 10 -11 m. Find: a) The force of attraction. b) The orbital period. Constants: e = 1. 6 x 10 -19 C me = 9. 11 x 10 -31 kg e a) Use Coulomb’s Law, and remember that q is the magnitude of the charge only. b) Use F=ma, and then circular motion. Answers: a) 5. 8 x 10 -7 N, b) 3. 5 x 10 -17 s

Example 2: A small plastic sphere has a known charge of +6. 5 µC. It is attracted to a second small sphere 3. 5 cm away with a force of 0. 023 N. Find: a) The charge on the second sphere. b) The number of excess electrons on the second sphere. a) - = 4. 8 x 10 -10 C negative! b) = 3. 0 x 109 electrons

Homework • Textbook pg. 142 #1 -6

Unit 7: Electrostatics Lesson 2: The Principle of Superposition

The Principle of Superposition The resultant force on any one particle equals the vector sum of all forces on it. Example 1: a) Find the net force on the – 5. 0 C charge. b) Where should that -5. 0 C charge be placed so that the net force is zero? 2. 0 C a) F = F 1 – F 2 + Q 2 -1. 0 C -5. 0 C _ _ 2 m 2 m Q 1 Q 5 _ F 2 F 1

b) 2. 0 C + Q 2 -1. 0 C -5. 0 C _ _ x 2 m Q 1 Q 5 _ F 2 F 1

Example 2: Three +20 µC charges are placed at the corners of an equilateral triangle of side 1. 0 m. Find the force exerted on one charge by the other two. y + F 1 F + + x 30 o F 2 F 1 = F 2 = k(20 x 10 -6)2 1. 02 = 3. 6 N F = 2 x F 1 x = 2 x F 1 cos 30 = 6. 2 N

Brain Break!

Q Fc Q (a) + Q + (b) Fa + db Fb 2 m θ 5 m + (c) Example 3: (Try it!) Find the force on the top right charge. (Q = 40 µC) Q X : Fx = Fa + Fbcosθ 2 2 = k. Q + k. Q cosθ d a 2 d b 2 = k. Q 2 ( 12 + 1 cos 21. 8) 5 29 = 1. 04 N Y : Fy = Fc + Fbsinθ = k. Q 2( 1 2 + 1 sin 21. 8) 2 29 = 3. 78 N F db =√(22 + 52) = √ 29 θ = tan-1(2/5) = 21. 8 o F = √(1. 0422 + 3. 7822) = 3. 9 N θ = tan-1(1. 04/3. 78) = 15 o 3. 9 N, 15 o right of the vertical

Homework • Textbook pg. 142 #7, 8, and 10