Alta Physics Chapters 16 - 19 Electrical Energy, Electric Fields & DC Circuits.

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Presentation transcript:

Alta Physics Chapters Electrical Energy, Electric Fields & DC Circuits

Alta Physics Other Forms of Stored Energy: Chemical Energy Stored in the Chemical Bonds that make up a substance Often released by combustion (burning) Released as kinetic energy Heat Light Sound *** Demonstration ***

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

Alta Physics 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.

Alta Physics 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

Alta Physics 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 C)(-15 x C)/(1.5 m) 2 = -0.6 N

Alta Physics 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

Alta Physics 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 C)/(0.3 m) 2 = N/C

Alta Physics 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

Alta Physics 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 C 1 V = 1 Joule/1 Coloumb Work = Volts * Charge = Vq

Alta Physics 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 C) = 7 x J

Alta Physics 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

Alta Physics Sample Problem What charge is stored when a 0.5 F capacitor is attached to a 9 volt source? Solution q = CV = (0.5 F)(9 V) = 4.5 Coloumbs

Alta Physics Capacitance To calculate the capacitance of a plate capacitator C = Kε 0 A/d where K = the dielectric constant ε 0 = the permitivity constant 8.85 x C 2 /Nm 2 A = the area of the plates in m 2 d = the distance between the plates in meters

Alta Physics Sample Problem What is the capacitance of a capacitor consisting of 2 plates, each having an area of 0.5 m 2, separated by 2 mm of mica? Solution C = Kε 0 A/d = (7)(8.85 x C 2 /Nm 2 )(0.5 m 2 )/(.002 m) = 1.55 x F = 1.55 nF

Alta Physics An Old Equation – with a twist Remember that the equation for the strength of an electric field is given by E = F/Q now we have V = W/Q where W = F x d so V/d = E or V = Ed

Alta Physics Electric Current 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.

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

Alta Physics 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

Alta Physics 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

Alta Physics 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

Alta Physics 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

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

Alta Physics 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 = ± 10% Or R = 25 MΩ ± 10%

Alta Physics Resistivity Spools or lengths of wire each have their own Resistance. Resistivity of these items can be calculated using the equation: R = ρL/A Where ρ is a constant, L is length, and A is cross sectional area of the wire.

Alta Physics Sample Problem Calculate the resistance of a spool of copper wire which is 20 m long and has a cross sectional area of 3.4 x m 2 ? Solution R = ρL/A= (1.68 x Ωm)(20 m)/(3.4 x m 2 ) = 1.14 x Ω

Alta Physics DC Circuits 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...

Alta Physics 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 V 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 A A = 0.75 A

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

Alta Physics 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

Alta Physics Problem Types Coloumb’s Law Electric Fields Voltage & Current Capacitors Circuits Batteries Ohm’s Law Power