electricity magnetism lecture 1: coulomb’s law - … lecture 01 - coulomb's...

Electricity & Magnetism Lecture 1: Coulomb’s Law

Today’s Concepts:

A) Coulomb’s Law

B) Superposi<on

Electricity & Magne<sm Lecture 1, Slide 1

If you haven't done Prelecture 1 yet, please do so later today

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.

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

q2 q1

q1

Like signs repel

Coulomb’s Law:

q2 q1

Opposite signs aMractq2 q1


Take two balloons and rub themboth with a piece of cloth. A8er you rub them they will:

A) A<ract each-‐otherB) Repel each-‐otherC) Either – it depends on the material of the cloth

Clicker Question: Balloons


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.

Coulomb’s Law


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).

q2q1

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

Our nota<on:

is the force by 1 on 2 (think “by-‐on ”)

is the unit vector that points from 1 to 2.

Clicker Question: Coulomb Force

What will the direc<on of the force be?

A) AMrac<ve B) Repulsive

k = 9 x 109 N m2 / C2

electron charge = 1.6 x 10−19 Coulombs

NA = 6.02 x 1023

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.


Clicker Question: Coulomb Force

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)

k = 9 x 109 N m2 / C2

electron charge = 1.6 x 10−19 Coulombs

NA = 6.02 x 1023

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.


CheckPoint: Forces on Two Charges


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. .”

q1

q2

q3

q4

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:

F2,1

F3,1

F4,1

F1

Superposition


F2,1

F3,1

F4,1

F1

What happens to Force on q1 if its sign is changed?

A) |F1| increases

B) |F1| remains the same

C) |F1| decreases

D) Need more informaKon to determine

Clicker Question: Superposition

q1

q2

q3

q4


The direc<on of all forces changes by 180o – the magnitudes stay the same:

F2,1

F3,1

F4,1

F1

F2,1

F3,1

F4,1

F1

F2,1

F3,1

F4,1

F1

q1

q2

q3

q4

F2,1

F3,1

F4,1

F1

q1

q2

q3

q4



qq

−Q +Q

+Q+Q

Case 1 Case 2

Compare the magnitude of the net force on q in the two cases.

A) |F1 | > |F2|

B) |F1 | = |F2|

C) |F1 |< |F2|

D) Depends on sign of q

F1F2 = 0

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.

CheckPoint: Compare Forces

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?

A) leftward B) none C) rightward

x

y

3qq

q

q

Q


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

What is the direc<on of horizontal force on Q?

A) Fx > 0 B) Fx = 0 C) Fx < 0

x

y

3q

q qq

Q


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?

A) Fy > 0 B) Fy = 0 C) Fy < 0

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.

x

y

3q

q qq

Q


electricity magnetism lecture 1: coulomb’s law - … lecture 01 - coulomb's...

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