Electricity and Magnetism

Electric Fields and Electric Charge Electric Charges Exert forces that can attract and repel each other even when they are not in direct contact Unit for a charge = Coulomb (named after Charles Augustin de Coulomb)

In 1911, Robert Millikan performed experiments (using oil drops) and made two important observations: –The smallest value for the charge on an oil drop is x C. –All other charge values are whole numbered multiples of x C

Therefore, overall charge on an object is represented by the equation: Q = Ne Where: –Q = amount of charge in coulombs –N = the total number of electrons in either deficit or excess –e = elementary charge (1.602 x C) (electron (-e) = x C; proton (e) = x C) For our purposes, we will only use e =1.60x C

Ex. 1. How many electrons have been removed from a positively charged pith-ball electroscope if it has a charge of 7.5x C? Given: Q=7.5x C e=1.6x C N=? Q = N e 7.5x = N (1.6x ) 7.5x = N (1.6x ) N = 4.7x10 8 Therefore, the number of electrons removed was 4.7x10 8 electrons.

Electric Fields The forces between electrical charges is due to a field of force Coulomb’s Law: (where k is a constant, Q’s are the charges of object 1 and 2, and d is the distance) Good news: No problems will be given in this course using this formula But beware in grade 12…

Electric Current Objects can develop charges. These charges can move from one object to another and be distributed over a conductor. The flow of electric charge is called an electric current (I).

–I is the current in amperes (A) –Q is the charge in Coulombs (C) …remember Q = Ne –t is the time in seconds Therefore 1 A = 1 C/s

Example 2. Calculate the current in an electric toaster if it takes 9.0x10 2 C of charge to toast 2 slices of bread in 1.5min. Given: t=1.5min t= 1.5minx60s/min =90s Q= 9.0x10 2 C I=? Therefore, the current in the toaster is 10 A.

Ex. 3. A metal-leaf electroscope with 1.25x10 10 excess positive charges is grounded and discharges completely in 0.50s. Calculate the average current in the grounding wire. Given: N=1.25x10 10 e=1.6x C t=0.50s I=? Q=Ne Q=(1.25x10 10 )(1.6x ) Q=2.0x10 -9 C =2.0x10 -9 C 0.50s =4.0x10 -9 A Therefore, the average current in the grounding wire is 4.0x10 -9 A. We need Q

Measurement of Electric Current An ammeter measures the amount of electric current in a circuit. It is connected directly into the path of moving charges (i.e. connected in series). Ammeters are excellent conductors. In direct current the current flows in a single direction from the power supply to the load. It does not increase or decrease in magnitude. (i.e. battery) In alternating current the current periodically reverses direction in the circuit and the amount of current varies continuously. (i.e. wall socket)

Electric Potential Difference (V) Similar to gravitational potential energy, an electric charge has an amount of electrical potential energy because of the electric field set up by the power supply For gravitational potential energy to increase, work must be done. In circuits, work is done by the power supply to increase the electrical potential energy of each coulomb. As the charge goes through the load, its energy decreases. As the charge goes through the power supply, its energy increases.

The electrical potential energy for each coulomb of charge in a circuit is call the electric potential difference or voltage. V = E / Q or sometimes seen as V = W / Q –V is the voltage in volts (V) –Q is the charge in Coulombs (C)…remember Q = Ne –W or E is the amount of work that must be done or energy required to increase the electric potential

Therefore 1 V = 1 J/C or in words, one volt is the electric potential difference between two points if one joule of work is required to move one coulomb of charge between the two points. A voltmeter is used to measure electric potential difference. Voltmeters are extremely poor conductors (high resistance). A voltmeter must be connected across a source or the load in a circuit (i.e. in parallel)

Unit Equations Electrical Power: Electrical Potential Difference: Current: Electrical Energy:

Calculating the Electrical Energy Lost or Work Done V = E/Q I = Q/t Therefore E = VQ and Q = It So ∆E = (V)(I)(∆t) E can be described as the electrical energy lost by a current (I) through a potential difference (V), for a time interval t.

Ex. 4. A 12-V car battery supplies 1.0x10 3 C of charge to the starting motor. How much energy is used to start the car? Given: V= 12V Q= 1.0x10 3 C E= ? E= Q V E= (1.0x10 3 C)(12V) E= 1.2x10 4 J Therefore, the amount of energy used to start the car is 1.2x10 4 J.

Ex. 5. If a current of 10.0A takes 300s to boil a kettle of water requiring 3.6x10 5 J of energy, what is the potential difference (voltage) across the kettle? Given: I=10.0A t=300s E=3.6x10 5 J V=? E=(V)(I)(t) V= E (I)(t) V= (3.6x10 5 J) (10A)(300s) V= 1.2x10 2 V Therefore, there is a potential difference of 1.2x10 2 V across the kettle.

Your night light uses a 7W light that draws about 0.06 A of current. How much charge passes through this bulb in 8 hours.