L01 - Ch 21 Electrostatics The study of the properties of...

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Unformatted text preview: Ch 21 Electrostatics The study of the properties of stationary charges Properties of Charge • Fundamental Property of Matter • Two types of charge positive negative • Like charges repel • Unlike charges attract • Charge is quantized (distinct fundamental units) e = 1.602 × 10-19 C Coulomb (C) is the SI unit of charge • Law of Conservation of Charge: The total charge of any isolated system is a constant Charge Conservation 238 92 U→ Th + He 234 90 4 2 There are 92 protons on either side of this nuclear reaction equation. Charge Conservation Particle Physics γ →e +e − + The decay of a neutral photon into and electron and a positron (positively charge electron) in GREEN !!! Method of generating static charge • Process of inducing a positive charge on a conducting sphere. • Contact or induction on isolated (insulated) conductors Material Behaviors • Insulator - large effort required to move charges through material – Examples: wood, plastic, rubber, glass, dry air, vacuum • Conductor - very little effort required to move charges through material – Examples: most metals, ionic solutions, plasmas • Semi-conductor - intermediate effort required to move charges through material – Examples: germanium, silicon Charge and Force Ιnsulators Conductor and Inuslator Coulomb’s Law (Force Law between charges) • Charles Auguste Coulomb - 1785 • Force between two point charges is proportional to each charge • Force between two point charges is inversely proportional to the distance separating them • Force between charges obey the Law of Superposition (vector addition) Coulomb’s Law (point Charges) r q1 q2 ˆ F =k r 2 r k = 8.99 × 10 N − m / C 9 2 2 1 k= 4 πε 0 ε 0 = 8.85 × 10 −12 C 2 / N − m 2 like charges unlike charges A positive 1.0 μC charge is located at the origin and a -0.3 μC charge is located at x = 2.0 cm. What is the force on the negative charge? Example 1 F =k q1 q2 r 9 2 1.0 μC 0.0 cm 2 −6 -0.3 μC 2.0 cm −6 N − m 1.0 × 10 C −0.3 × 10 C F = 8.99 × 10 2 2 C ( 0.02m ) F = 6.74 N r $ F = −6.74 Ni A positive 0.1 μC charge is located at the origin, a +0.2 μC charge is located at (0.0 cm, 1.5 cm), and a -0.2 μC charge is located at (1.0 cm, 0.0 cm). What is the force on the negative charge? 3 0.2 μC (0.0 cm, 1.5 cm) Example 2 1 α 2 -0.2 μC (1.0 cm, 0.0 cm) r q1 q2 ˆ F =k r 2 r 0.1 μC (0 cm. 0cm) Determine the force between the negative charge and each positive charge. 3 0.2 μC (0.0 cm, 1.5 cm) 2 0.1 × 10 −6 C − 0.2 × 10 −6 C r N −m ˆ F21 = 8.99 × 109 i 2 2 C ( 0.01m ) 1 0.1 μC (0 cm. 0cm) 2 -0.2 μC (1.0 cm, 0.0 cm) r $ F21 = − 1.80 Ni The force between charges 1 and 2. r q1 q2 ˆ F21 = k i 2 r21 3 0.2 μC (0.0 cm, 1.5 cm) 1 α 2 -0.2 μC (1.0 cm, 0.0 cm) 0.1 μC (0 cm. 0cm) The force between charges 3 and 2. 2 0.2 × 10 −6 C − 0.2 × 10 −6 C r 9 N −m F23 = 8.99 × 10 2 2 C2 ( 0.01m ) + ( 0.015m ) r F23 = 111N . 3 0.2 μC (0.0 cm, 1.5 cm) 1 α 2 -0.2 μC (1.0 cm, 0.0 cm) 0.1 μC (0 cm. 0cm) Determine the components of each force. r $ F23 = − F23 cos α i + F23 sin α$ j 0.01m $ . i + 111N r . F23 = −111N 0.015m (0.01m )2 + (0.015m )2 (0.01m )2 + (0.015m )2 r $ $ F23 = − 0.62Ni + 0.92Nj $ j 3 0.2 μC (0.0 cm, 1.5 cm) 1 α 2 -0.2 μC (1.0 cm, 0.0 cm) 0.1 μC (0 cm. 0cm) Determine the sum of the components of the forces. r $ $ $ F = −1.80Ni − 0.62Ni + 0.92Nj Determine the resultant force $ $ F = − 2.42 Ni + 0.92 Nj r F = 2.59 N @ 159 o from +x axis Sample prob 21-3 a) What is force between spheres b) Ground sphere A, then what is the force ? ...
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