Electric Charges & Electric Forces Chapter 20
Electric Charges & Electric Forces Chapter 20 Structure of atom Mass (kg) Charge (Coulombs) Proton 1.673X10-27 +1.60X10-19 Neutron 1.675X10-27 0 = + e Electron 9.11 X 10-31 0 mass -1.60X10-19 = - e
Example: A rod has a charge of -4.0 C. How many electrons must be removed so that the charge becomes +2.0 C? 1 C = 1 X 10-6 C Solution The total amount of electrons that need to be removed is 6.0 C or 6.0 X 10-6 C, and 1 e- has a charge of -1.60 X 10-19 C 6.0 X 10-6 C (1 e-/1.60 X 10-19 C) = =3.75 X 10 13 electrons
Law of Conservation of Electrical Charge Electrons may be transferred from one object to another, but they will never disappear or appear from nowhere. Electron abundant = negative charge Electron deficient = positive charge
Electrically charged objects... Exert a force on each other Unlike charges attract Like charges repel
Electrical Conductors In most metal atoms, the outer electrons are not held tightly by the nucleus these electrons can freely roam/move. Generally, good thermal conductors are good electrical conductors. Examples: silver, copper, aluminum
Electrical Insulators In materials where the outer electrons are held tightly by the nucleus these electrons cannot freely roam/move. Generally, good thermal insulators are good electrical insulators. In a power line, charge flows much more easily through hundred of kilometers of metal wire than through the few centimeters of insulating material that separates the wire from the supporting tower. In a common appliance cord, charges will flow through several meters of wire to the appliance, and then through its electrical network, and then back through the return wire rather than flow directly across from one wire to the other through the tiny thickness of rubber insulation. In general, electrons follow path of least resistance.
Charging by Friction It is possible to transfer electric charge from one object to another. Walk across carpeted floor with wool socks and you can build up an excess of charge (either on you or the carpet) and therefore, become charged. Can use friction to remove electrons from fur and deposit onto rod. Fur becomes positively charged. Why? Rod becomes negatively charged. Why?
Charging by Contact The process of giving one object a net electric charge by placing it in contact w/another object that is already charged is known as charging by contact. (Conduction)
Charging by Induction The process of giving one object a net electric charge without touching the object to a second charged object is called charging by induction. Polarization
Temporarily Charging by Induction An object may become polarized (opposite ends) while a charged object is brought near. But, the overall charge on this piece of plastic (for example) is neutral. When the charged rod goes away, the electrons rearrange and disperse evenly. Object is still neutral. Polarization
Question: A negatively charged rod is brought near a neutral metal sphere. Which of the following is true? A. There is an attractive force between the rod and sphere. B. There is a repulsive force between the rod and sphere. C. There is no electric force between the rod and sphere. Slide 20-6
Answer: A negatively charged rod is brought near a neutral metal sphere. Which of the following is true? A.There is an attractive force between the rod and sphere ( the neutral sphere is polarized temporarily). B. There is a repulsive force between the rod and sphere. C. There is no electric force between the rod and sphere.
Visualizing Charge Charges on an insulator do not move. Charges on a conductor adjust until there is no net force on any charge. We call this electrostatic equilibrium.
Coulomb s Law When oppositely charged objects are near each other, there is a force experienced by each attracting them to one another. True for the same charge objects too except the force is pushing the objects apart and a repelling force exists. According to Newton s 3 rd law, these forces are equal and opposite.
Coulomb s Law The electrostatic force that stationary charged objects exert on each other depends on the amount of charge (q) on the objects and the distance (r) between them. Since these objects are so small compared to the distance between them, they may be regarded as point charges. F = k (q 1 q 2 )/r 2 F = the amount of force that particle #1 exerts on #2 and vice versa (same force!) (Unit: Newtons) k = proportionality constant = 8.99 X 10 9 N m 2 /C 2 q 1 = amount of charge of particle #1 (Unit: Coulombs) q 2 = amount of charge of particle #2 (Unit: Coulombs) r = the distance between q 1 and q 2 (Unit: meters)
Checking Understanding A small, positive charge is placed at the black dot. In which case is the force on the small, positive charge the largest? Slide 20-24
Answer: C A small, positive charge is placed at the black dot. In which case is the force on the small, positive charge the largest? C C Slide 20-25
Slide 20-26 Checking Understanding A small, positive charge is placed at the black dot. In which case is the force on the small, positive charge the smallest?
Answer: D A small, positive charge is placed at the black dot. In which case is the force on the small, positive charge the smallest? D D
Checking Understanding All charges in the diagrams below are of equal magnitude. In each case, a small, positive charge is placed at the black dot. In which case is the force on the small, positive charge the largest?
Slide 20-30 Answer: C All charges in the diagrams below are of equal magnitude. In each case, a small, positive charge is placed at the black dot. In which case is the force on the small, positive charge the largest? C
You can have more than two charged particles interacting. To solve for the net force recall that net force is the sum of all the forces. Solve for the x and y components if the forces are not in one plane. Then add the x-components, add the y- components, and solve for the resultant, using the Pythagorean theorem. Don t forget direction! Next slides show two examples
What s the net electrostatic force on q 1? q 1 is being attracted (pulled toward) both q 2 and q 3.
Solution F 12 = (kq 1 q 2 )/(r 122 ) = plug in values F 12 = 2.7 N F 13 = (kq 1 q 3 )/(r 132 ) = plug in values F 13 = 8.4 N Since F 12 points in the x direction, and F 13 points in the +x direction: F = -F 12 + F 13 = -2.7N + 8.4N = F = +5.7 N The net force is 5.7 N pointing to the right.
What s the net electrostatic force on q 1?
Solution: F 12 = (kq 1 q 2 )/(r 122 ) = plug in values F 12 = 9.6 N directed 73 o above +x axis, having both an x-component and y-component F 12x : cos 73 o = F 12x / F 12 F 12x =(F 12 )(cos 73 o )=9.6N(cos 73 o ) = +2.8N F 12y : sin 73 o = F 12y / F 12 F 12y =(F 12 )(sin 73 o )=9.6N(sin 73 o ) = +9.2N F 13 = (kq 1 q 3 )/(r 132 ) = plug in values F 13 = 18 N directed along +x axis F = [(F 12x + F 13 ) 2 + (F 12y ) 2 ] = F = [(2.8N+ 18N) 2 + (9.2N) 2 ] = 23 N = tan -1 (Fy/Fx) = tan -1 (9.2N/20.8N) = 24 o