Electrostatic Forces Atoms: protons (p+) bound in nucleus electrons (e-) freer to move about e- and p+ have equal amounts but opposite charge When objects rub together, electrons may be exchanged Objects that have more e- than p+ are negative Objects with fewer e- than p+ are positive Charge can not be created or destroyed, only transferred. Charge is “quantized” - there is a smallest amount that can be isolated (the amount on one e- or p+).

Electrostatic Forces Unit of Charge (q): the Coulomb (C) 1 Coulomb = 6.24 x 1018 e- |Charge on one qe- | = 1.60 x 10-19 C

Electrostatic Forces Unit of Charge: the Coulomb (C) 1 Coulomb = 6.24 x 1018 e- Charge on one e- = 1.60 x 10-19 C Example Problem: A penny has a charge of -1.00 x 10 -15 C. How many More electrons than protons does it have? Answer: #e- = q x e-/q -1.00 x 10-15 C x 6.24 x 1018 e-/C = 6.24 x 10 3 excess e-

Electrostatic Forces Conductors and insulators Conductors: e- free to move about Conductors: mostly metals Insulators: e- not free to move about Insulators: generally non-metals

Electrostatic Forces Remember the general form for Gravitational force? G = 6.67 x 10-11 Nm2/kg2 Well electrostatic force is another “force-at-a-distance”, and Has a very similar form…. k = 9.0 x 109 Nm2/C2

Electrostatic Forces k = 9.0 x 109 Nm2/C2 Use this equation to find the magnitude of the mutual force on any two point charges. Don’t bother plugging in the signs of the charges. If the point charges have opposite signs, the force is attraction. If they have the same sign the force is repulsion. Each object experiences the same amount of force, even if one has more charge or mass – remember Newton’s third law!

Electrostatic Forces Lets compare the gravitational and electrostatic forces on a proton and electron located 1.0 m apart…. me- = 9.11 x 10-31 kg qe- = -1.60 x 10-19 C mp+ = 1.67 x 10-27 kg qp+ = +1.60 x 10-19 C G = 6.67 x 10-11 Nm2/kg2 k = 8.99 x 109 Nm2/C2 Fg = 1.01 x 10-67 N (attraction) Fe = 2.30 x 10-28 N (also attraction)

Voltage and Current Voltage = change in potential energy per charge Unit: 1 volt = 1 joule/coulomb Current = rate of charge flow Unit: 1 ampere = 1 coulomb/second

Georg Ohm (1789 - 1854) came up with the fact that for most materials, the current through the material is proportional to the voltage across the material, at least for some range of voltages and temperatures. The proportionality constant is known as Resistance (R) . The equation that is commonly referred to as Ohm’s law is generally written: The Unit of resistance is appropriately called an Ohm. The symbol for the Ohm is W. Materials that obey Ohm’s “Law” are said to be “Ohmic”

Two factors that determine the resistance of an object: The material’s “resistivity” (r) (good conductors have low resistivity, insulators have high resistivity) The size and shape of the object (longer, narrower paths have higher resistance) These two factors establish the following relationship: resistivity A L length resistance cross section area

Example question: The four resistors to the right are all made of the same material, and have dimensions as shown. Put them in order from lowest to highest resistance. d L 2d 2L I. II. III. IV. Remember: Smaller L, smaller R Bigger A, smaller R

Example question: Put them in order from lowest to highest resistance. d L 2d 2L I. II. III. IV. Answer: Least R: IV. (small L, big A) 2nd R: III. (A = pr2, bigger factor) 3rd R: I. Most R: II. (big L, small A) Remember: Smaller L, smaller R Bigger A, smaller R

And now back to Ohm’s Law... Example Problem: A 1.0 m long wire is connected to a 12 volt battery. 20 coulombs of charge flows through it in 10. seconds. A) What is the current in the wire? I = Q/t I = 20 C/10 s = 2 C/s = 2 A

And now back to Ohm’s Law... Example Problem: A 1.0 m long wire is connected to a 12 volt battery. 20 coulombs of charge flows through it in 10. seconds. B) What is the resistance of the wire? V = IR R = V/I R = 12 V / 2 A R = 6 W

And now back to Ohm’s Law... Example Problem: A 1.0 m long wire is connected to a 12 volt battery. 20 coulombs of charge flows through it in 10. seconds. C) What would the resistance of the wire be if it were 2.5 m long? R is proportional to L L = 2.5 Lo R = 2.5 Ro R = 2.5 x (6W) = 15 W

P = VI What do you get when you multiply Voltage x Current? Answer: Think definitions: Voltage = energy/charge Current = charge/time P = VI

Kirchhoff’s Rules Junction Rule Loop Rule The total current flowing into a junction must equal the total current flowing out of the junction. (Due to conservation of charge) Loop Rule The sum of the potential changes around any closed loop must equal zero. (Due to conservation of energy) 2 V 7 A V = 24 V 4 A 14 V 3 A 8 V

Series and Parallel Circuits When devices are connected in an electric circuits, they can be connected in “series” or in “parallel” with other devices. When devices are series, any current that goes through one device must also go through the other devices. For example: Series Connection 1 2 A B The devices, numbered “1” and “2” in the diagram above, are connected in series. If an electron (or even conventional positive current) needs to move from point A to point B, it must go through both device 1 and device 2. Everything that goes through one must also go through the other.

Series and Parallel Circuits When devices are connected in an electric circuits, they can be connected in “series” or in “parallel” with other devices. When devices are parallel, the current goes through one device only. For example: Parallel Connection 1 2 A B The devices, numbered “1” and “2” in the diagram above, are connected in parallel. If an electron (or even conventional positive current) needs to move from point A to point B, it must go through only one device, not both. Some current goes through one, some through the other.

Resistors in Series Resistors in Parallel 1 2 A B If devices 1 and 2 are resistors, think of the series connection as a longer resistor. Longer resistors have greater resistance. The total resistance of a combination of resistors in series is greater than any of the individual resistances. Resistors in Parallel 1 2 A B If devices 1 and 2 are resistors, think of the parallel connection as a wider resistor. Wider resistors have lower resistance. The total resistance of a combination of resistors in paralle is smaller than any of the individual resistances.

Series Circuits V R2 R1 R3 The schematic circuit diagram to the right shows three resistors (R) connected in series with a source of potential difference (V). Rules for a simple series circuit…. (in sentence form) 1) The total equivalent resistance of resistors in series is equal to the sum of the individual resistances. 2) The sum of the voltage drops across each of the resistors is equal to the total voltage of the power supply. 3) The same amount of current flows through all the resistors. 4) The total power converted by the three resistors is equal to the sum of the individual powers

Rules for a simple series circuit…. (in equation form) Series Circuits V R2 R1 R3 The schematic circuit diagram to the right shows three resistors (R) connected in series with a source of potential difference (V). Rules for a simple series circuit…. (in equation form) Req = R1 + R2 + R3 +... V = V1 + V2 + V3 + ... I = I1 = I2 = I3 = ... P = P1 + P2 + P3 + ...

Parallel Circuits R3 V R1 R2 The schematic circuit diagram to the right shows three resistors (R) connected in parallel with a source of potential difference (V). Rules for a simple parallel circuit…. (in sentence form) 1) The reciprocal of the total equivalent resistance of resistors in series is equal to the sum of the reciprocals of the individual resistances. 2) All of the resistors have the same voltage drop across them. 3) The sum of the currents through all the parallel resistors is equal to the total current supplied by the voltage source. 4) The total power converted by the three resistors is equal to the sum of the individual powers.

Parallel Circuits R3 V R1 R2 The schematic circuit diagram to the right shows three resistors (R) connected in parallel with a source of potential difference (V). Rules for a simple parallel circuit…. (in equation form) V = V1 = V2 = V3 = …. I = I1 + I2 + I3 +…. P = P1 + P2 + P3 + ….

Series Circuits: Example Problem V = 24 V R2 = 4 W R1 = 2 W R3 = 6 W Req = R1 + R2 + R3 +… V = V1 + V2 + V3 + … I = I1 = I2 = I3 = … P = P1 + P2 + P3 + ... Fill in Given Use Req = R1 + R2 + R3 +… to find the total equivalent resistance. 1 V I R P 2 3 T Use V = IR to find the total current 2 4 6 24 4 8 12 2 8 16 24 48 I = I1 = I2 = I3 = … Use V = IR to find the individual voltages Use P = VI to find all the powers 2 12

Parallel Circuits: Example Problem V = V1 = V2 = V3 = ... I = I1 + I2 + I3 + ... P = P1 + P2 + P3 = ... V = 50 V R1 = 20 W R2 = 25 W R3 = 100 W Fill in Given V = V1 = V2 = V3 = ... Use to find the total equivalent resistance. 1 V I R P 2 3 T 20 25 100 50 50 2.5 2 0.5 5 125 100 25 250 Use V = IR to find the individual currents Use P = VI to find all the powers 10

Combination Circuits Now R2 is in parallel with R4, 3. V R1 R2 R4 R6 R5 The circuit to the right is a complicated combination circuit. The resistors aren’t all in series or all in parallel. To analyze a combination circuit, first look for pairs (or more) of resistors which are in series or parallel. In this example R3 and R4 are in series, while R5 and R6 are in parallel. Use the series and parallel rules to replace the pairs with “equivalent” resistors. V R1 R2, 3, 4 R5, 6 V R1, 2, 3, 4, 5, 6 V R1 R2 R4, 3 R5, 6 Now R2 is in parallel with R4, 3. Now R2, 3, 4 is in series with R1 and R5, 6.

Combination Circuits R3 = 12 W V = 120 V R1 = 19 W R2 = 48 W R4 = 12 W R6 = 30 W R5 = 6 W Example: Find the equivalent resistance of the six resistors in the circuit at right. 1/R5,6 = 1/R5 + 1/R6 = 1/30W + 1/6 W R5,6 = 5 W R3,4 = R3 + R4 = 12 W + 12 W R3,4 = 24 W 1/R2,3,4 = 1/R2 + 1/R3,4 = 1/48W + 1/24W R2,3,4 = 16 W R1,2,3,4,5,6 = R1 + R2,3,4 + R5,6 R1,2,3,4,5,6 = 19 W + 16 W + 5 W R1,2,3,4,5,6 = 40 W

Combination Circuits V = 30 V R1 = 18 W R2 = 9 W R3 = 6 W R4 = 12 W This circuit is neither a simple series circuit or a simple parallel circuit - it is a combination of the two. The resistors marked R1, R2, and R3 are in parallel with each other, their combination is in series with R4. Use the parallel rule for R1, R2, and R3 Then use the series rule for R1, 2, 3 and R4 Req = R1, 2, 3 + R4 = 15 W R1, 2, 3 = 3 W

Combination Circuits V = 30 V R1 = 18 W R2 = 9 W R3 = 6 W R4 = 12 W We found the total equivalent resistance to be 15 W, now let’s try to fill in the whole table… Since we know the VT and REQ we can find IT. We can see that all the current goes through R4... 1 V I R P 2 3 4 T So we can now get V4 18 9 6 12 30 6 .33 2 4 6 48 Because R4 is in series with the combination of R1, 2, 3 the voltages must add to RT. .67 1 24 2 Now you can complete the table 2 15 60

Combination Circuits: Hints When the current reaches point A, it must split to the right or the left. If R1 has a bigger resistance than R2, most of the current will go through R2. If R1 = 2 x R2, then twice as much current will go through R2 as compared to R1. I1 + I2 = I 2 x I1 = I2 I1 = (1/3)I I2 = (2/3)I R2 R1 A B Answer: If R1 = 5 x R2, then five times as much current will go through R2 as compared to R1. I1 + I2 = 30 A 5 x I1 = I2 I1 = (1/6)(30 A) = 5 A I2 = (5/6)(30 A) = 25 A R1 = 50 W R2= 10 W A B I = 30 A Example Problem: How much of the 30 A of current goes through each resistor in the diagram at right?

Combination Circuits Example Problem: The circuit to the right represents a battery and Four identical light bulbs connected in a combination circuit. Q: Put them in order from brightest to dimmest: Brightest: R4 (all current goes through it) 2nd R1 (gets 2/3 of total current) 3rd & 4th R2 and R3 (get 1/3 of total current) Now the bulb represented by R3 is unscrewed from its socket. R2 R1 R3 R4 R2 R1 R3 R4 Q: What happens to the brightness of the other three bulbs? R2 goes out (no current through that branch) R4 gets dimmer (fewer current paths = more total resistance in circuit = less total current = less through R4) R1 gets brighter (current through R4 decreases, voltage across R4 decreases, voltage across R1 increases = it gets brighter)

Combination Circuits R3 V R1 R2 R4 R6 R5 R7 R8 R9 R10 The circuit to the right consists of ten identical resistors. Put the resistors in order from the greatest amount of current to the least current. Answer: Most current: R1 (only resistor which all the current passes through) 2nd: R2 (gets 2/3 of the total current) 3rd: R5 and R6 (get 1/2 of the total current) 4th: R3 and R4 (get 1/3 of the total current) Last: R7, R8, R9, R10 (get 1/4 of total current)