STATIC ELECTRICITY.

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STATIC ELECTRICITY

A. Definition- Study of the forces between charges at rest.
B. Micro structure of matter (atomic structure) 1. Atom- 3 Parts a. Protons-mass 1 amu, charge +1, location nucleus b. Neutrons-mass 1 amu, charge 0,

c. Electrons- very small mass, charge
-1, location outside the nucleus C. Charged Objects 1. Ion-electrically charged particles that result from gain or loss of electrons a. Lose electrons- positive ions b. Gain electrons-negative ions

2. Interaction of charged particles
a. Like charges are repelled by an electrical force that acts on each particle in opposite directions + Fe Fe

b. Unlike charges attract each other by an electrical force that also acts in opposite directions.
+ - Fe Fe Neutral charges are attracted to both positive and negative but are NEVER repelled

D.Electroscope-device used to measure charge.
KNOB LEAVES

1. charging an electroscope by conduction means a charged object is touched to the top of the electroscope. When the scope is touched by negatively charged object the excess electrons move from the negatively charged object down the metal rod of the scope and to the leaves of the eletroscope . This causes the leaves to have like charge (negative) and move apart (diverge).

Negatively charged rod
- Electrons traveling Down the rod and Through the electroscope Making the scope negatively charged e- e- e-

b. When a positively charged rod is touched to an electroscope the electrons move from the scope to the rod leaving the knob and leaves of the scope positive and the leaves repel each other

+ + + Positively charged rod Electrons are attracted to the positively charged rod and leave the scope and enter the rod. Leaving both the knob and leaves of the electroscope positively charged. e- e- e-

2. To charge an electroscope by induction simply bring a charged rod near the top of the scope- do not touch it. a. When a negatively charged object is brought near a neutral scope the electrons in the top are repelled and move down into the leaves. This results in the leaves being negatively charged and the top being positively charged. Since the leaves are both negative they repel each other and move apart

- ++++ e- e- e- e- e-

b. When a positively charged rod is brought near a neutral electroscope electrons from the leaves are attracted to the top, resulting in the top having a negative charge and the leaves having a positive charge. Since the leaves are both positive they repel each other and move apart

+ ------ e- e e-

F. Conservation of charge
E. Grounding 1. Occurs when the earth or other large conductor accepts or donates a large number of electrons without significantly affecting its own electrical state. 2. Grounding results in an object having a net charge of zero also known as discharging. F. Conservation of charge Law In an isolated system (charge carriers cannot enter or leave) net charge is constant.

2. Example: 2 identical conducting spheres A and B are separated by some distance.
Sphere A has a net charge of -16 Sphere B has a net charge of +4 total charge If A and B are brought into contact electrons will move from A to B until both have a charge of -6, but the net charge on the system remains at -12

G. Measurement of Charge 1
G. Measurement of Charge 1. The smallest isolated negative charge in nature is that of an electron and for positive charge the proton. 2. An elementary charge is symbolized by e- (negative charge) or e+(positive charge like a proton) 3. Unit of charge is the Coulomb a. 1coulomb= 6.25X1018 elementary charges b. 1e- has a charge of -1.6X10-19C (elementary charge)

c. 1 proton has a charge of +1.6X10 -19C See reference tables d. Net charge depends on excess or deficiency of electrons therefore a net charge is always a whole number multiple of the charge of one electron and can be expressed that way. 2e = 2(1.6 x10-19C)= 3.2 X 10-19C

Coulomb’s Law 1. Measures the force two charged objects exert on each other when separated by a distance. 2. Fe=kq1q2 r2 Where k = constant = 8.99X109 N-m2/C2 F = force in Newtons q1 & q2= charge in coulombs r= separation distance in meters

b., The electrostatic force of each object is directed along the line joining the
two objects and is equal in magnitude but opposite in direction for the two objects

Examples: What is the force between 2 relatively small charged objects that carry charges of 0.20C and -0.30C, if the distance between them is 1m? Fe=kq1q2 Fe= 8.99X109N-m2/C2(.2C)(-.3C) r (1m)2 Fe= -539,400,000N

A positive charge of 6. 0 X10-6 C is 0
A positive charge of 6.0 X10-6 C is 0.30m from a second positive charge of 3.0X10-6C. Calculate the force between the charges. Fe=kq1q2 Fe= 8.99X109N-m2/C2(6X10 -6 C)(3X10-6C) r (0.30m)2 Fe=+1.8N

Electric Fields A. Produced by charged objects B
Electric Fields A. Produced by charged objects B. Field lines are normal perpendicular to the surface of the charge C. Point away from positve charges Point toward negative charges D. Field intensity inside the charge is zero___

+ -

E.Calculating field strength E= Fe q E= electric field intensity, N/C Fe= electrostatic force, N q.= charge, C Field direction is the same as the force for a positive charge and 180 degrees opposite for a negative charge. Field strength indicated by closeness of lines.

Example: A positive test charge of 4 X 10-5 C is placed in an electric field . The force on it is 0.60N acting at 10 degrees. What is the magnitude and direction of the electric field? E=Fe E= 0.6N E=15,000N/C q x10-5C

G. Parallel Plates The field that exists between two oppositely charged parallel plates is uniform as long as their separation distance is small

Electric Potential 1. The work needed to bring a unit positive charge toward a positive object. 2. Conversely the work done to pull to oppositely charged objects apart + WORK +++++ ++++ Positively charged object Test charge --- ---- WORK +++ WORK

Potential Difference A. The change in energy of a positive test charge as it is moved closer to a positively charged object B. The volt 1. Unit of potential difference 2. 1volt=1joule/1coulomb 3. Definition: In an electric field when 1 Joule of work is required to move a one coulomb of charge between two points

4. V= W q V=potential difference, volts, V W= work, joules, J
4. V= W q V=potential difference, volts, V W= work, joules, J .q =charge, Coulombs, C

Example: What is the potential difference between two points in an electric field if 10J of work is required to move a 2C charge between them? V= W q V= 10J 2C V=5J/C = 5volts

What is the work required to move 5 coulombs of charge through a potential difference of 10V? V= W q 10V= W 5C W=50J

6. Electronvolt (eV) a. Definition-Amount of energy required to move one electron through a potential difference of 1volt. It is a unit of work/energy just like the joule. 1eV = 1.60 X10-19 joule

III. Current electricity A
III. Current electricity A. Definition- Flow of electrons from high potential (V) to low potential (V) B. Required conditions a. A potential difference supplied by a battery or generator. b. A complete or closed conducting path connecting regions of high potential to low potential (like a wire)

2. Unit- a. 1 amp of current is measured when one coulomb of charge flows past a point each second 3. Calculated I = ∆q t I = current in ampere,A q= charge in coulombs t= time in seconds

Example: What is the current if 15C of charge pass a point in 3 seconds? I=q t I=15C/3s I= 5C/s or 5A

What is the amount of charge passing through a lamp in 10 seconds if it draws 0.5Amps of current? I=q t 0.50A= q 10s q=5C 4. An Ammeter is used to measure current

D. Resistance, R 1. Determines the amount of current that can be produced by a conductor. 2. Causes there to be a potential difference between the ends of a conductor when a current is passed through it Resistance depends on the length, cross-sectional area, and resistivity of the conductor and temperature

4. Resistivity- Physical property of material to resist or oppose the movement of charge through a material 5. Calculating Resistance R=ρL A R = resistance, in ohms (Ω)=V/A ρ = (greek letter rho) resistivity, ohm-meter found on ref. tables page4 L= length, meters A= cross-sectional area, m2

Example: Calculate the resistance at 20oC of an aluminum wire that is 0.200m long and has a cross-sectional area of 1X10-3 square meter. R=ρL A R= 2.82x10-8 Ω-m(0.200m) 1x10-3m2 R= 5.64X10-6 Ω

6. Temperature a. Resistance is directly related to temperature b
6. Temperature a. Resistance is directly related to temperature b. Low temp, low resistance and vice versa

E. Ohm’s Law. R= V I R= resistance, ohms V= potential difference, volts I=Current, amps

AMMETER(CURRENT) VOLTMETER(POTENTIAL DIFF)
Graph Symbols AMMETER(CURRENT) VOLTMETER(POTENTIAL DIFF) P D O I T F E F N E T R I E A N L C E V CURRENT, I (AMPS) A V RESISTANCE, Ω

Example: If the current across a wire is 2A and the potential difference across the wire is 10V, R= V I R=10V 2A R=5 Ω

A potential difference of 12V is applied across a circuit which has a 4 ohm resistance. What is the magnitude of the current in the circuit? R= V I 4Ω =12V I= 3 amps, 3A, 3C/s

F. Power P=VI= I2R = V2/R P =power, watts (J/s) V = potential difference, volts I = current, amp R= resistance, Ω

Example: A lamp operates at 10volts and draws a current of 0
Example: A lamp operates at 10volts and draws a current of 0.5amps for 60 seconds. What power is developed in the lamp? P=VI= I2R = V2/R P=VI P=10V(0.5A) P= 5watts

While operating at 120V, an electric toaster has a resistance of 15 ohms. What is the power used by the toaster? P=VI= I2R = V2/R P= V2/R P= (120V)2/15 Ω P= 960 watts

G. Work-Electrical Energy
W= Pt=VIt= I2Rt = V2t/R P =power, watts V = potential difference, volts I = current, amps R= resistance, t = time, s W = work , J

Examples: A toaster dissipates 1500 watts in 90 seconds. What is the amount of electrical energy used by the toaster? W= Pt=VIt= I2Rt = V2t/R W=Pt W=1500watts(90s) W=135,000J

An iron has a current of 10Amps when 120V of potential difference is applied for 60s. What is the total energy dissipated during the 60seconds? W= Pt=VIt= I2Rt = V2t/R W=VIt W=120V(10A)(60s) W=72,000J

H. Conservation of charge and energy 1
H. Conservation of charge and energy 1. Conservation of charge Kirchhoff’s first rule: For any point in a circuit, the total current arriving at a point must equal the total current leaving the point. 2. Conservation of energy Kirchhoff’s second rule: The algebraic sum of all the voltage drops and applied voltage (battery) around a circuit is zero

Example: 2A 3A ? 6A Current into the juncture: 3A+6A=9A
Current leaving the juncture: 2A+? Since current in must equal current out ?=7A

Series Circuit 1. Only one path for the current to follow Sketch 2. Current Same through each resistor b. IT = I1= I2 = I3 ….. c. Read by an ammeter d. Connected in series Symbols- VOLTAGE (BATTERY)

Resistance a. The total resistance is equal to the sum of the components (resistors) b. RT = R1 + R2 + R3 …. Symbol See previous notes Voltage-(potential drops)- VT= V1 + V2+V3 +… REMEMBER: R=V I

Make a sketch of the circuit
A 10 ohm and a 20 ohm resister are connected in series with a 60V battery. Make a sketch of the circuit 10Ω 20Ω 60V

b. What is the total resistance of the circuit. RT = R1 + R2 + R3 …
b. What is the total resistance of the circuit? RT = R1 + R2 + R3 …. RT = 10Ω +20Ω RT= 30Ω c. What is the total current in the circuit? R=V/I 30Ω = 60V I I= 2 amps Since it is a series circuit the current is the same everywhere

d,. What is the voltage drop across the 10 ohm resistor
d,. What is the voltage drop across the 10 ohm resistor? R=V/I 10Ω = V 2A V= 20V e. What is the voltage drop across the 20 ohm resistor? 20Ω = V V= 40V

f. What is the current through the 10 ohm resistor?
2A-CURRENT IS THE SAME EVERYWHERE IN A SERIES g. What is the current through the 20 ohm resistor?

Parallel Circuit 1. Circuit in which the current can flow through various paths

Current a. The current is divided among the different branches. b
Current a. The current is divided among the different branches. b. The total current is equal to the source (battery) c. IT = I1+ I2 + I3 ….. Potential Drops (voltage) a. The potential difference across each branch (resistor) is equal to the total (battery voltage) b. VT = V1= V2 = V3

Resistance 1 = RT R1 R2 R3 REMEMBER: R=V/I APPLIES TO BOTH SERIES AND PARALLEL CIRCUITS.