1. Introduction to Electricity Electric charges come in two varieties. We have named these positive and negative. To be mathematically consistent all of electricity is defined from the perspective of positive charges. When there are two or more charges in a region, they will experience a force. The electric force is a vector. To determine the force on one charge in the vicinity of one or more other charges, you must first determine the force on that one charge from each of the other charges in the region and add them as individual vectors. The force can also be determined if the charge is in a known electric field. The force will be the product of the charge and the strength of the field.

2. Charges repel if they have the same sign ( + + or - - ), and they attract if the signs are opposite ( + - ). Coulomb’s Law is the equation that is used to determine the force between a pair of charges: F = k q 1 q 2 k = Coulomb’s Constant k = 8.987 x 10 9 Nm 2 /C 2 (on your equation sheet) q 1 and q 2 are the two charges (in units of Coulombs, C) r is the distance between the charges r2r2 What is the force between a proton and an electron if they are separated by 1.50 x 10 -10 m? F = k q 1 q 2 r2r2 F = (8.987 x 10 9 Nm 2 /C 2 ) (1.602 x 10 -19 C)(1.602 x 10 -19 C) (1.50 x 10 -10 m) 2 F = 1.025 x 10 -8 N = 1.03 x 10 -8 N

3. Michael Faraday developed the idea that there is a field that surrounds each charge and that this field is what causes the force on other charges in the region. The electric field (E) is the identified as the force per unit charge from one or more charges in a region. E = k q r2r2 E = F / q The electric field can be written in the same form as Coulomb’s Law: The electric field is also a vector. The direction of the electric field is the direction the force would be if a POSITIVE charge were the field. The electric field is always away from positive charges and toward negative charges. This is not a UNIFORM field. It varies with distance

4. What is the electric field 0.250 m from a 15.0  C (15.0 x 10 -6 C) charge? E = k q r2r2 E = (8.987 x 10 9 Nm 2 /C 2 ) (15.0 x 10 -6 C) (0.250 m) 2 E = 2.157 x 10 6 N/C = 2.16 x 10 6 N/C What is the force on an electron at that location? E = F / qF = q E = 1.602 x 10 -19 C (2.157 x 10 6 N/C) F = 3.45 x 10 -13 N There is energy stored in electric fields much like gravitational potential energy in gravitational fields. This electric potential energy (like all energy) is NOT a vector.

5. Electric Fields around Simple Arrangements Point Charges + -

6. + - A Dipole Field

7. + + Like Charges Two negative charges will have the same field pattern except the arrows would be on the other end (toward the charges).

8. When the position of a mass determines the force on the mass, there is a potential energy involved ( W =  E = F d ). Moving in a field of force means there is a change in potential energy. The electric potential energy ( U ) is determined with an equation similar to Coulomb’s Law and will have units of Joules, but it is NOT A VECTOR. In the force equation ( Coulomb’s Law ), we ignored the signs of the charges to find the magnitude and used the signs to determine direction. In the energy equation, SIGNS MUST BE INCLUDED! A negative potential energy means that it takes energy to escape the effects of the forces (zero energy is at a distance of ).

9. The equation for electric potential energy is: U = k q 1 q 2 r The amount of potential energy per unit charge is called ELECTRIC POTENTIAL ( V ) which has units of VOLTS. 1 Volt = 1 Joule per Coulomb The equation for electric potential from one charge is: V = U / q OR V = k q r The equation for electric potential from multiple charges is: V =  k q r

10. Two charges are placed 0.400 m apart. The one on the left has a charge of + 245  C, and the one on the right has a charge of – 125  C. What is the electric field midway between the charges? E 1 = k q 1 r12r12 = 8.987 x 10 9 Nm 2 /C 2 ( 245 x 10 -6 C ) ( 0.200 m ) 2 = 5.50454 x 10 7 N/C E 2 = 8.987 x 10 9 Nm 2 /C 2 ( 125 x 10 -6 C ) ( 0.200 m ) 2 = 2.8084x 10 7 N/C The direction of the field from both charges will be to the right, so they add together. E = 8.31 x 10 7 N/C Now determine the electric potential of a point midway between the two charges.

11. V 1 = k q 1 r1r1 = 8.987 x 10 9 Nm 2 /C 2 ( + 245 x 10 -6 C ) ( 0.200 m ) = 1.1009 x 10 7 V V 2 = 8.987 x 10 9 Nm 2 /C 2 ( - 125 x 10 -6 C ) ( 0.200 m ) = - 5.6169 x 10 6 V V = 1.1009 x 10 7 V + ( - 5.6169 x 10 6 V ) = 5.39 x 10 6 V Two charges of the same magnitude and opposite signs are placed near each other. The one on the left has a positive charge, and the one on the right has a negative charge. What is the electric field midway between the charges? Both fields will have the same magnitude and be in the same direction (to the right). E = 2 K q / r 2 What is the electric potential midway between the charges? Both potentials will have the same magnitude but opposite signs; they cancel out! V = 0

12. It takes work to move a charge from one potential ( V a ) to another ( V b ). For a charge, this is the same as having a change in potential energy. W =  U = q V a – q V b = q V ab A change in electrical potential is called a POTENTIAL DIFFERENCE. A potential difference of 1 volt will take or add 1 Joule of energy to 1 Coulomb of charge.

13. When a potential difference (V ab ) is placed across two parallel plates (usually from a battery), a UNIFORM field is created between the plates. This means that the electric field has the same value at any point between the two plates. V ab One plate will be positively charged and one will be negatively charged. + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - The field lines will go from the top ( + ) towards the bottom ( - ). V ab = Ed d = the distance between the plates

14. How much energy would an electron gain when moving across a potential difference of 12.0 V between two parallel plates? V = U / q  U = q V ab = e V ab  U = 1.602 x 10 -19 C (12.0 V) = 1.92 x 10 -18 J What is the electric field between the two plates if they are separated by 0.0600 m? V ab = Ed 12.0 V = E (0.0600 m) E = 12.0 V / (0.0600 m) = 200 V/m = 200 N/C What is the force on the electron between the two plates? F = qEF = (1.602 x 10 -19 C) (200 N/C) F = 3.204 x 10 -17 N = 3.20 x 10 -17 N How much work is done to move the electron 0.0600 m against that force? W = FdW = (3.204 x 10 -17 N) (0.0600 m) W = 1.92 x 10 -18 J The answer could also be: 12.0 eV!

15. ELECTRICITY: Circuits Current ( I ) is measured in Coulombs per second. One Coulomb per second is called an Ampere or Amp. The unit of the Ampere is (A) The “force” that causes current is called the electromotive force which is a potential difference (voltage) and not really a force. Batteries are our primary source of voltage in circuits. When there is something to cause a force on charges and those charges are free to move, they will begin to flow. The movement of charge is called current. Symbol for battery: + - I

16. Resistance means that energy is being lost (usually in the form of heat). This is electrical potential energy (not the kinetic energy of the charges). Resistance can be used to direct current since it will have a greater value through paths with less resistance. A device that is intentionally used to direct current through its resistance is called a resistor. Symbols for resistors: When a current crosses a resistor, the potential of the current drops in the direction of the current. As current travels it nearly always encounters some electrical friction called RESISTANCE ( R ). The unit of resistance is the Ohm (  )

17. The following equation shows the relationship between these three values: V = I R As current ( I ) encounters RESISTANCE ( R ) it loses potential ( or voltage, V ). If the resistance is constant, this is OHM’s LAW If a path is followed around a circuit (of wires and resistors) the total potential lost will equal the total created by the battery. There are two devices used to measure current and potential difference: The ammeter measures current in a wire. The voltmeter measures the potential difference between two points in a wire. A V

18. The following is a simple circuit with a battery and a resistor. An ammeter to measure the current and a voltmeter to measure the potential difference across the resistor are included. V The battery will drive a current counterclockwise around the circuit. A R The voltage from the battery will be lost through the resistor: V – I R = 0 If there is more than one resistor, then you can use the concept of equivalent resistance. We will look at two common arrangements. When the equivalent resistance is determined, the single resistance (R E ) replaces the entire arrangement. The voltage of the battery will be entirely used by the circuit.

19. Resistors in series are arranged one right after another along a single wire of the circuit. R1R1 R3R3 R2R2 R E = R 1 + R 2 + R 3 + … Resistors in parallel are arranged on separate wires from the main wire of the circuit. R1R1 R3R3 R2R2 1 / R E = 1 / R 1 + 1 / R 2 + 1 / R 3 + … OR: R E = [ 1 / R 1 + 1 / R 2 + 1 / R 3 + … ] -1

20. What is the equivalent resistance of a 25 , 33 , and 15  resistor if they are in parallel? 1 / R E = 1 / R 1 + 1 / R 2 + 1 / R 3 R E = [ 1 / 25  + 1 / 33  + 1 / 15  ] -1 = 7.3  What is the potential difference across each one of the resistors if hooked to a 12 V battery? 25  33  15  Each will have 12 V across it. What is the current through each resistor? V = I R I = V / R I 1 = 12 V / 25  = 0.48 A I 2 = 12 V / 33  = 0.36 A I 3 = 12 V / 15  = 0.80 A What would be the current through the single equivalent resistor? I E = 12 V / 7.3  = 1.6 A 0.48 A + 0.36 A + 0.80 A = 1.6 A

21. The following three resistors are hooked up in series to a 8.0 V battery: 4.0 , 10.0 , and 6.0  What is the current through each resistor? 4.0  6.0  10.0  R E = R 1 + R 2 + R 3 R E = 4.0  + 10.0  + 6.0  R E = 20.0  V = I R I = V / R I E = 8.0 V / 20.0  = 0.40 A What is the potential difference across each resistor? V = I R V 1 = 0.40 A ( 4.0  ) = 1.6 V V 2 = 0.40 A ( 10.0  ) = 4.0 V V 3 = 0.40 A ( 6.0  ) = 2.4 V

22. Resistors in series will all have the same current through them but the potential difference for each depends on the resistance of each. Resistors in parallel will all have the same potential difference across them but the current for each depends on the resistance of each. Resistors Current Potential Difference Parallel Series Depends on Resistors Same for all Resistors Depends on Resistors Same for all Resistors

23. Resistors take the energy from the circuit and expel it as heat. The rate of this energy transfer is the power ( P ) of the resistor. P = I V = V 2 / R = I 2 R A light bulb is rated as: 75 W / 120 V What is the resistance of the bulb? P = V 2 / RR = V 2 / P R = ( 120 V ) 2 / 75 W = 192  = 190  What is the power output of the above bulb if connected to a 40.0 V potential difference. P = V 2 / R = ( 40 V ) 2 / 192  P = 8.333 W = 8.33 W

24. Kirchhoff’s rules for circuits: Loop Rule: The sum of all potential differences around any closed path in a circuit must equal zero (like we had with our single loop earlier). Resistors create ( - ) V ab ( I R ) in the direction of current. Batteries create ( + ) V ab in the direction of current. The loop rule is a conservation of energy rule. The net change in energy should be zero if you are back to the starting point. Point Rule: The sum of all currents into and out of a given point must equal zero.  I in =  I out The point rule is a conservation of charge rule. charge cannot be created or destroyed at any point in a circuit.

25. Electricity Practice Problems What is the current drawn from the battery in the circuit below? V ab = 12.0 V R 1 = 2.00  R 2 = 3.00  R 3 = 4.00  R1R1 R2R2 R3R3 V ab V ab = I R E R E = 2.00  + 3.00  + 4.00  = 9.00  12.0 V = ( I ) 9.00  I = 1.33 A A voltmeter is attached as shown. What should the voltmeter read? V V = I R = ( 1.3333 A ) 3.00  = 4.00 V What is the total power output of the entire circuit? P = I V = 1.3333 A ( 12.0 V ) = 16.0 W P = V 2 / R E = ( 12.0 V ) 2 / 9.00  = 16.0 W P = I 2 R E = ( 1.3333 A ) 2 ( 9.00  ) = 16.0 W

26. The ammeter in the following circuit reads 1.45 A. Determine the potential difference (VOLTAGE) of the battery. R 1 = 1.00  R 2 = 2.00  R 3 = 4.00  V ab = ? R1R1 R2R2 R3R3 V 1 = ( 1.45 A ) 1.00  = 1.45 V V ab = 1.45 V + 2.90 V + 5.80 V = 10.15 V V 2 = ( 1.45 A ) 2.00  = 2.90 V V 3 = ( 1.45 A ) 4.00  = 5.80 V

27. What is the current drawn from the battery? V ab = 10.0 V R 1 = 1.00  R 2 = 2.00  R 3 = 2.00  R 4 = 3.00  R1R1 R2R2 R3R3 R4R4 V ab R E = ? 1 / R E = 1 / 2.00  + 1 / 2.00  + 1 / 3.00  1 / R E = 1.3333   R E = 0.750  1.00  + 0.750  = 1.750  I 1 = 10.0 V / 1.750  = 5.71 A

28. Two charges are arranged 60.0 cm apart as shown. The charge on the left is + 3.00  C, and the charge on the right is – 3.00  C. What is the electric field midway between the two points? 60.0 cm E =  k q r2r2 = 2 k q r2r2 E = 2 (8.987 x 10 9 Nm 2 /C 2 ) 3 x 10 -6 C (0.30 m) 2 = 5.99 x 10 5 N/C What is the electric potential at the same point? V = k q r  = 0 What is the force on an electron at that point? F = q E = 1.602 x 10 -19 C ( 5.99 x 10 5 N/C ) = 9.60 x 10 -14 N

29. Three charges of equal magnitude but different sign are set up at the corners of an isosceles triangle as shown. What is the direction of the force on the charge at the top. The signs are indicated. + + - The charge will be repelled by the lower left charge. The charge will be attracted to the lower right charge. The distances and magnitudes are the same for each pair meaning they will produce the same forces; the upward component of one will be equal and opposite to the downward component of the other. The resultant will be to the right.

30. What would be the kinetic energy of the alpha particle if it moved 1.00 cm from one plate to another plate if there is a potential difference of 24.0 V between the plates? V = U / q  K =  U = q V ab = 2e V ab  K = 2 ( 1.602 x 10 -19 C ) (24.0 V) = 7.69 x 10 -18 J 48.0 eV What is the averge speed of an electron if it is orbiting a proton at a distance of 1.50 x 10 -10 m? F = k q 1 q 2 r2r2 F = (8.987 x 10 9 Nm 2 /C 2 ) (1.602 x 10 -19 C)(1.602 x 10 -19 C) (1.50 x 10 -10 m) 2 F = 1.025 x 10 -8 N = mv 2 / r v 2 = ( 1.50 x 10 -10 m ) 1.025 x 10 -8 N / ( 9.109 x 10 -31 kg ) v = 1.30 x 10 6 m/s