herriman high honors physics chapters 16 - 18 electrical energy, electric fields & dc circuits

Herriman High Honors Physics

Chapters 16 - 18

Electrical Energy, Electric Fields

& DC Circuits


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 ***


Electric Charge and Electric Field

Static Electricity – Unmoving charge Two types

Positive – lack of electrons Negative – excess electrons

Like charges - Repel Opposite Charges - Attract


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.


Force Exerted by Charges Coulomb’s Law

F = kQ1Q2/r2

k = 9 x 109 N•m2/C2

Positive solution – repulsion Negative solution - attraction


Sample Problem

Two charges, Q1 = +10 µC, and Q2 = -15 µC, are separated by 1.5 meters.

What is the electrostatic force acting between them?

SolutionF = kQ1Q2/r2 =

(9 x 109 N•m2/C2)(+10 x 10-6 C)(-15 x 10-6 C)/(1.5 m)2

= -0.6 N

Practice AP. 566 #1 &

3


Electric Field Field – Affect that acts at a

distance, without contact Examples

Electric Field Gravitational Field

Electric Field Strength – E = F/q = kQ/r2


Sample Problem

Calculate the strength of an electric field at a point 30 cm

from a point charge Q = +3 µC

SolutionE = kQ/r2 =

(9 x 109 N•m2/C2)(+3 x 10-6 C)/(0.3 m)2

= 300000 N/C

Practice DP. 575 # 1 &

3


Chapter 17:Electrical Energy & Current

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


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


An Old Equation – with a twist

Remember that the equation for the strength of an electric field is given by

E = F/Qnow we have

V = W/Q where W = F x dso

V/d = E or V = Ed


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

Practice AP. 599 # 1 & 3


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


Sample Problem What charge is stored when a 0.5 F

capacitor is attached to a 9 volt source?

Solutionq = CV = (0.5 F)(9 V)

= 4.5 Coloumbs


Capacitance To calculate the capacitance of a

plate capacitatorC = Kε0A/d

where K = the dielectric constant

ε0 = the permitivity constant 8.85 x 10-12 C2/N•m2

A = the area of the plates in m2

d = the distance between the plates in meters


Sample Problem What is the capacitance of a capacitor

consisting of 2 plates, each having an area of 0.5 m2, separated by 2 mm of mica?

SolutionC = Kε0A/d

= (7)(8.85 x 10-12 C2/N•m2)(0.5 m2)/(.002 m)

= 1.55 x 10-9 F = 1.55 nF


Energy and Capacitors Energy stored in capacitors is

electric potential energy.

Where Q is the charge on one plate and ΔV is the voltage or potential difference

Sample Problem A capacitor connected to a 12 V battery holds

36 µC of charge on each plate. What is the capacitance of the capacitor and how much electrical potential energy is stored in the capacitor?

Solution


Practice BP. 607 #2 & 4


Electric Current & Resistance

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.


Electric Current Electron Flow is from – terminal to

+ terminal. Conventional Current is from +

terminal to – terminal.


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


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


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?SolutionR = V/I =

(1.5 V)/(0.3 A) = 5 Ω


Resistor Color Code Resistors are banded in order to

describe the amount of resistance they provide. Each resistor is banded with 4 stripes.

Band Represents1 First Digit2 Second

Digit3 Multiplier4 Tolerance


Bright Black 0

Boys Brown 1

Remember Red 2

Our Orange 3

Young Yellow 4

Girls Green 5

Become Blue 6

Very Violet 7

Good Grey 8

Wives White 9

Gold 5%

Silver 10%

None 20%

Resistor Color Code


Sample Problem

Calculate the resistance of a resistor which is banded with

the following colors: Red, Green, Blue, Silver.

SolutionRed = 2, Green = 5, Blue = 6 and Silver =

10% R = 25000000 ± 10%

OrR = 25 MΩ ± 10%


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. Practice D

P. 615 #1,3,& 5


Electric Power Power = Work/time In an Electical System P = QV/t So P = VI = I2R = V2/R


Sample Problem Calculate the resistance of a 40

Watt headlight which is designed to run on 12 Volts.

SolutionR = V2/P

R = (12 V)2/40 Watts = 36 Ω


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 10-6 m2?

SolutionR = ρL/A=

(1.68 x 10-8Ω•m)(20 m)/(3.4 x 10-6 m2) = 1.14 x 10-12 Ω


Chapter 18:DC Circuits

Batteries Connected in Series Increase Voltage

Et= E1 + E2 + E3. . . Produce the Same Current

It= I1 = I2 = I3. . . Batteries Connected in Parallel

Produce the Same VoltageEt= E1 = E2 = E3. . .

Increase CurrentIt= I1 + I2 + I3. . .


Sample Problem

Calculate the voltage and current when 3 batteries (1.5 V, 0.25 A are connected in

A) SeriesB) Parallel

Solutiona) Et= E1 + E2 + E3 =1.5 V + 1.5 V + 1.5 V = 4.5

VIt= I1 + I2 + I3= 0.25 A

b) Et= E1 = E2 = E3=1.5 VIt= I1 + I2 + I3=0.25 A + 0.25 A + 0.25 A = 0.75 A


DC Circuits Resistance in Series

Rt=R1+R2+R3. . . Resistance in Parallel

...1111

321 RRRRt


Sample ProblemCalculate the resistance when a 5 Ω, 6 Ω,

and 3 Ω resistor are connected in A) SeriesB) Parallel

Solution

a) Rt=R1+R2+R3 = 5 Ω+ 6 Ω+ 3 Ω = 14 Ωb)

Rt= 1.43 Ω

30

21

30

10

30

5

30

6

3

1

6

1

5

11111

321 RRRRt


In Class Practice Practice A, P. 650 1,3, & 5 Practice B, p. 655 2 & 4 Do Practice C, p. 659 1 & 2 on

board Do Practice D, p. 662 on board

herriman high honors physics chapters 16 - 18 electrical energy, electric fields & dc circuits

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