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Chapter 32 Electrostatics. Electric Charge and Electric Field Static Electricity – Unmoving charge Two types Positive – lack of electrons Negative – excess.

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Presentation on theme: "Chapter 32 Electrostatics. Electric Charge and Electric Field Static Electricity – Unmoving charge Two types Positive – lack of electrons Negative – excess."— Presentation transcript:

1 Chapter 32 Electrostatics

2 Electric Charge and Electric Field Static Electricity – Unmoving charge Two types Positive – lack of electrons Negative – excess electrons Like charges - Repel Opposite Charges - Attract

3 Electric Charges Charge can be induced by rubbing an object – View demonstrations Charge is detected using an electroscope. Charge can travel via a conductor. Poor conductors are insulators.

4 Force Exerted by Charges Coulomb’s Law F = kQ 1 Q 2 /r 2 k = 9 x 10 9 Nm 2 /C 2 Positive solution – repulsion Negative solution - attraction

5 Sample Problem Two charges, Q 1 = +10 µC, and Q 2 = -15 µC, are separated by 1.5 meters. What is the electrostatic force acting between them? Solution F = kQ 1 Q 2 /r 2 = (9 x 10 9 Nm 2 /C 2 )(+10 x 10 -6 C)(-15 x 10 -6 C)/(1.5 m) 2 = -0.6 N

6 Conductors and Insulators A good conductor transfers charge easily. A good insulator inhibits the transfer of charge. A good conductor is a poor insulator and a good insulator is a poor conductor.

7 Creation of Static Charges Two ways to create a static charge are: Charging by contact (friction) – when electrons are transferred from one object to another by touching. Induction – when a charge is transferred by bringing one object near another without actually touching

8 Chapter 33 Electric Fields and Potential

9 Electric Field Field – Affect that acts at a distance, without contact Examples Electric Field Gravitational Field Electric Field Strength – E = F/q = kQ/r 2

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11 Sample Problem Calculate the strength of an electric field at a point 30 cm from a point charge Q = +3 µC Solution E = kQ/r 2 = (9 x 10 9 Nm 2 /C 2 )(+3 x 10 -6 C)/(0.3 m) 2 = 300000 N/C

12 Electrical Energy Electrical Energy is generated from other forms of energy and transmitted over power lines and/or stored in batteries Vocabulary Voltage (V) Force in an electrical system; Volt = Work/Charge = W/q = Joule/Coloumb Current (I) Rate in an electrical system = Charge/time = q/t =Coloumb/sec = 1 Ampere

13 Energy in Electrical System Volts =Work/charge = V =W/q Work is measured in joules (the same as energy) Charge is measured in Coloumbs (C) The charge on an electron is 1.6 x 10 -19 C 1 V = 1 Joule/1 Coloumb Work = Volts * Charge = Vq

14 Sample Problem How much work is needed to move a 10 μC charge to a point where the potential is 70 V? W = Vq = (70 V)(10 x 10 -6 C) = 7 x 10 -4 J

15 Electrical Energy Storage Electrical Energy can be stored in two ways: Batteries Long term storage, even flow of charge Storage ability measured in Volts Capacitors Short term storage, releases charge all at once (boost in charge) Storage capacity measured in Farads (F) 1 Farad = 1 Coloumb/Volt Mathematically Charge = Capacitance * Voltage = q = CV

16 Chapter 34 Electric Current

17 Circuit – A continuous path connected between the terminals of a power source. Current – Flow of Charge I = ΔQ/Δt Current is measured in Coloumbs/Sec which is called an Ampere.

18 Electric Current Electron Flow is from – terminal to + terminal. Conventional Current is from + terminal to – terminal.

19 Sample Problem A steady current of 2.5 Amps passes through a wire for 4 minutes. How much charge passed through any point in the circuit? Solution Q = IΔt (2.5 C/s)(240 s) = 600 C

20 Ohm’s Law Resistance – how much the conductor slows down the flow of electrons through it. Resistance is measured in Ohms (Ω) Ohm’s law -In any Circuit: V = IR or R = V/I

21 Sample Problem A small flashlight bulb draws a current of 300 mA from a 1.5 V battery. What is the resistance of the bulb? Solution R = V/I = (1.5 V)/(0.3 A) = 5 Ω

22 Resistor Color Code Resistors are banded in order to describe the amount of resistance they provide. Each resistor is banded with 4 stripes. BandRepresents 1First Digit 2Second Digit 3Multiplier 4Tolerance

23 BrightBlack0 BoysBrown1 RememberRed2 OurOrange3 YoungYellow4 GirlsGreen5 BecomeBlue6 VeryViolet7 GoodGrey8 WivesWhite9 Gold5% Silver10% None20% Resistor Color Code

24 Sample Problem Calculate the resistance of a resistor which is banded with the following colors: Red, Green, Blue, Silver. Solution Red = 2, Green = 5, Blue = 6 and Silver = 10% R = 25000000 ± 10% Or R = 25 MΩ ± 10%

25 Electrical Power Electrical Power is measured in Watts. Power = current x voltage P = IV or P = I 2 R or P = V 2 /R Since Energy is Power x Time electrical energy is often measured in Kilowatthours or power x time.

26 Chapter 35 DC Circuits

27 Batteries Connected in Series Increase Voltage E t = E 1 + E 2 + E 3... Produce the Same Current I t = I 1 = I 2 = I 3... Batteries Connected in Parallel Produce the Same Voltage E t = E 1 = E 2 = E 3... Increase Current I t = I 1 + I 2 + I 3...

28 Sample Problem Calculate the voltage and current when 3 batteries (1.5 V, 0.25 A are connected in A) Series B) Parallel Solution a)E t = E 1 + E 2 + E 3 =1.5 V + 1.5 V + 1.5 V = 4.5 V I t = I 1 + I 2 + I 3 = 0.25 A b) E t = E 1 = E 2 = E 3 =1.5 V I t = I 1 + I 2 + I 3 =0.25 A + 0.25 A + 0.25 A = 0.75 A

29 DC Circuits Resistance in Series R t =R 1 +R 2 +R 3... Resistance in Parallel

30 Sample Problem Calculate the resistance when a 5 Ω, 6 Ω, and 3 Ω resistor are connected in A) Series B) Parallel Solution a)R t =R 1 +R 2 +R 3 = 5 Ω+ 6 Ω+ 3 Ω = 14 Ω b) R t = 1.43 Ω


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