Electricity & Magnetism
Lecture 1: Coulomb’s Law
Today’s Concepts:
A) Coulomb’s Law
B) Superposi<on
If you haven't done Prelecture 1 yet, please do so later today
Electricity & Magne<sm Lecture 1, Slide 1
Coulomb’s Law:
The force on a charge due to another charge is propor<onal to the product of the charges and inversely propor<onal to the separa<on squared. q2
q1
r
The force is always parallel to a line connec<ng the charges, but the direc<on depends on the signs of the charges:
q2
q1
q2
q1
q2
q1
Opposite signs aMract
Like signs repel
Electricity & Magne<sm Lecture 1, Slide 2
Clicker Question: Balloons
Take two balloons and rub them
both with a piece of cloth. A8er you rub them they will:
A) A<ract each-­‐other
B) Repel each-­‐other
C) Either – it depends on the material of the cloth
Electricity & Magne<sm Lecture 1, Slide 3
Balloons
If the same thing is done to both balloons they will acquire the same sign charge. They will repel!
Text
+
+
+
+
+
+
+
+
We can only be sure about this if the balloons are made out of the same material.
Electricity & Magne<sm Lecture 1, Slide 4
Coulomb’s Law
Our nota<on:
is the force by 1 on 2 (think “by-­‐on ”) is the unit vector that points from 1 to 2. Examples: If the charges have the same sign, the force by charge 1 on charge 2 would be in the direc<on of r12 (to the right).
q1
q2
If the charges have opposite sign, the force by charge 1 on charge 2 would be opposite the direc<on of r12 (leT).
q1
q2
Electricity & Magne<sm Lecture 1, Slide 5
Clicker Question: Coulomb Force
Two paperclips are separated by 10 meters. Then you remove 1 electron from each atom on the first paperclip and place it on the second one. k = 9 x 109 N m2 / C2
electron charge = 1.6 x 10−19 Coulombs
NA = 6.02 x 1023
What will the direc<on of the force be? A) AMrac<ve B) Repulsive
Electricity & Magne<sm Lecture 1, Slide 6
Clicker Question: Coulomb Force
Two paperclips are separated by 10 meters. Then you remove 1 electron from each atom on the first paperclip and place it on the second one. k = 9 x 109 N m2 / C2
electron charge = 1.6 x 10−19 Coulombs
NA = 6.02 x 1023
Which weight is closest to the approximate force between those paperclips (recall that weight = mg, g = 9.8 m/s2)?
A) Paperclip (1 g x g)
B) Text book (1 kg x g)
C) Truck (104 kg x g)
D) AircraE carrier (108 kg x g)
E) Mt. Everest (1014 kg x g)
Electricity & Magne<sm Lecture 1, Slide 7
CheckPoint: Forces on Two Charges
Electricity & Magne<sm Lecture 1, Slide 8
CheckPoint Results: Forces on Two Charges
“The bigger charge will push the smaller charge harder than the smaller will push the larger. “
“According to Newton's Third Law, the charges exert equal and opposite forces on each other. “
“The forces of two objects are always equal and opposite. both equasions involve the charge of both parOcles in the same manner, therefore they are the same. .” Electricity & Magne<sm Lecture 1, Slide 9
Superposition
If there are more than two charges present, the total force on any given charge is just the vector sum of the forces due to each of the other charges:
q2
F2,1
F4,1
F1
q1
F3,1
F1
F3,1
F2,1
F4,1
q4
q3
Electricity & Magne<sm Lecture 1, Slide 10
Clicker Question: Superposition
What happens to Force on q1 if its sign is changed?
A) |F1| increases
B) |F1| remains the same
C) |F1| decreases
q2
D) Need more informaKon to determine
q1
q4
q3
Electricity & Magne<sm Lecture 1, Slide 11
The direc<on of all forces changes by 180o – the magnitudes stay the same:
q2
q2
F4,1
F4,1
F1
q1
F2,1
F1
F3,1
q3
F2,1
F2,1
F3,1
q4
q3
F1
q1
q4
F2,1
F3,1
F3,1
F4,1
F1
F4,1
Electricity & Magne<sm Lecture 1, Slide 12
CheckPoint: Compare Forces
Compare the magnitude of the net force on q in the two cases.
A) |F1 | > |F2|
+Q
+Q
q
q
B) |F1 | = |F2|
C) |F1 |< |F2| F2 = 0
F1
−Q
+Q
Case 1
Case 2
D) Depends on sign of q
In case one, both of the forces are working in the same direcOon (the downward direcOon) resulOng in a greater net force as opposed to case two where the repelling forces are working in opposite direcOons in turn canceling each other out Both cases have a charged placed equidistant from charges of the same magnitudes so no ma<er charge q has the forces at this instant are the same. There is no way to tell if the forces on the center charge are repulsive or a<racOve without knowing the sign of the center charge. Electricity & Magne<sm Lecture 1, Slide 13
Force from Four Charges
Four charged par<cles are placed as shown. A par<cle with charge +Q is placed 1/2 way between the middle +q on the leT and the +3q on the right
What is the direc<on of horizontal force on Q? y
x
q
q
Q
3q
A) leftward B) none C) rightward
q
Electricity & Magne<sm Lecture 1, Slide 15
CheckPoint Results: Force from Four Charges
Four charged par<cles are placed on a circular ring with radius 3 m as shown below. A par<cle with charge Q is placed in the center of the ring
y
3q
What is the direc<on of horizontal force on Q?
x
Q
A) Fx > 0 B) Fx = 0 C) Fx < 0
q
q
q
Electricity & Magne<sm Lecture 1, Slide 16
Clicker Question: Force from Four Charges
Four charged par<cles are placed on a circular ring with radius 3 m as shown below. A par<cle with charge Q is placed in the center of the ring
What is the ver&cal force on Q?
y
3q
A) Fy > 0 B) Fy = 0 C) Fy < 0
x
Q
q
q
q
Fy would equal 0 if all 3 of the bo<om charges were on top of each other, but because they are pulling at an angle, they will not equal the force of the 3q charge that is pulling only in the +Y direcOon. The y-­‐direcOon components also cancel each other out since there is a charge or force of 3Q poinOng both upward and downward. The two q charges on the side also have an x component so their Fy is not as great thus the sum of the force from the bo<om charges in smaller than the top charge. The forces are repulsive so the charge will have a negaOve Fy. Electricity & Magne<sm Lecture 1, Slide 17