Physics 1308: General Physics II - Professor Jodi Cooley

Charles-Augustin de Coulomb 1736-1806

Welcome Back! Physics 1308

Coulomb’s Law

Physics 1308: General Physics II - Professor Jodi Cooley

Announcements• Assignments for Tuesday, August 28th:

- Reading: Chapter 22.1 - 22.2

- Watch Video: https://youtu.be/tZq79CsdtTw — Lecture 3 - The Electric Field

• Homework 1 Assigned - due before class on Tuesday, August 28th.

- Numerical answers to WileyPLUS problems and answers to all WileyPLUS questions should be entered into WileyPLUS

- You must also generate a set of written solutions to ALL your assigned questions and solutions to ALL your assigned problems for each assignment. One of those problems or questions will be randomly selected and graded each week for methodology.

- Written solutions will need to follow the homework policy and will be graded according to that policy. We expect that you will produce good-quality homework solutions! Points will be lost for not adhering to the homework guidelines.

• Dr. Cooley will be out of the office the week of August 27th. Her office hours are cancelled for that week. Please email her if you have any immediate concerns or needs.

Physics 1308: General Physics II - Professor Jodi Cooley

• Dr. Cooley’s office hours will be in FOSC 151 on Wednesdays from 3 - 4 pm and Fridays from 1 - 2 pm or by appointment.

• Ms. Jasmine Liu’s office hours will be in a location FOSC 153 on Fridays from 10-11 am and Mondays from 2 - 3 pm.

Physics 1308: General Physics II - Professor Jodi Cooley

Please join us for the PHYSICS Department Kick-Off Event

Drop in from 6:30 – 7:30 pm Tuesday Aug 28th Heroy Hall 153

Pizza & Ice cream Meet the Physics faculty and undergraduate Physics majors

Learn about opportunities for undergraduate research Get free advice about majoring or minoring in Physics Find out about the Society of Physics Students (SPS)

Physics 1308: General Physics II - Professor Jodi Cooley

Quiz Time

Physics 1308: General Physics II - Professor Jodi Cooley

Review Question 1An initially electrically neutral conducting sphere is placed on an insulating stand. A negatively-charged glass rod is brought near, but does not touch the sphere. Without moving the rod, a wire is then attached to the sphere that connects it to earth ground. The rod and wire are then removed simultaneously. What is the final charge on the sphere?

A) negative

B) positive

C) neutral

D) it has a 50/50 chance of having either a positive or negative charge

Physics 1308: General Physics II - Professor Jodi Cooley

Matter and Size

• An atom is about a tenth of a nanometer in size (10-10m, called 1 “Angstrom”)

• We in the macroscopic (human-sized) world are about 1 meter in size

• The macroscopic world is 10 billion times bigger than the atoms that compose it. To us, they are like idealized geometric “points” whose individual size can be neglected, but whose sum explains us.

• There are approximately 1027 atoms in our body... more atoms in our body than there are stars in the visible Universe (estimated to be at least 100 billion trillion, or 1023).

Physics 1308: General Physics II - Professor Jodi Cooley

Charges and Point Charges

• Charge distributions can be complex . . .

• . . . but as far as we know, all charge distributions are made from point-like charges.

• Coulomb's Law describes the electric force between two point charges

• It will be our archetype for solving problems involving charge distributions.

Physics 1308: General Physics II - Professor Jodi Cooley

Key Equations and ConstantsCoulomb’s Law:

F =1

4⇡✏0

|q1||q2|r2

~F = Gm1m2

r2r

Newton’s Equation:

F =1

4⇡✏0

|q1||q2|r2

~F = Gm1m2

r2r

k =1

4⇡✏0= 8.99⇥ 109 N ·m2/C2

✏0 = 8.99⇥ 10�12 N ·m2/C2

Electrostatic Constant:

Permittivity Constant:

Physics 1308: General Physics II - Professor Jodi Cooley

Problem 1A current of 0.300 A through your chest can send your heart into fibrillation, ruining the normal rhythm of heartbeat and distributing the flow of blood (and thus oxygen) to your brain. If that current persists for 1.34 minutes, how many conduction electrons pass through your chest? Hint: 1 A = 1 C/s

|q| = (0.300 C/s)(1.34 min)(60 s/min) = 24.12 C

The number of charges is given by the formula

Solution:

Find the total charge in 1.34 minutes.

q = ne

n =q

e=

24.12 C

1.602⇥ 10�19 C= 1.51⇥ 1020

n = 1.51 x 1020

Physics 1308: General Physics II - Professor Jodi Cooley

Physics 1308: General Physics II - Professor Jodi Cooley

Physics 1308: General Physics II - Professor Jodi Cooley

Problem 2: Ions Near a Cell Membrane• Membrane potential is the difference in electric

potential between the interior and exterior of a biological cell (typically from -40 mV to - 80 mV).

• Barrier serves as an insulator and diffusion barrier to movement of ions.

• Ion transporters or ion pump proteins actively push ions across the membranes.

• Ion pump proteins and ion channels are electrically equivalent to a set of batteries and resistors inserted into the membrane (more on that later).

• Do the ions feel an electrostatic force?

By Synaptidude, CC BY 3.0, https://commons.wikimedia.org/w/index.php?curid=21460910

Physics 1308: General Physics II - Professor Jodi Cooley

Problem 2: Ions Near a Cell MembraneLet’s make a simple model. What is the electrostatic force that the potassium ion (charge = +1e, m = 6.5 x 10-26 kg) exerts on the sodium ion (charge +1e, m = 3.8 x 10-26 kg)?

L = 2.8 x 10-9 m

Solution:

The electrostatic force is given by Coulomb’s Law:

We need to take note of two things. First, we need to calculate the numerical value of the charge for both ions. Second, notice that this is a two dimensional problem. We will have to take account of both the x- and y- directions.

~F =1

4⇡✏0

|q1||q2|r2

r

Physics 1308: General Physics II - Professor Jodi Cooley

Recall that the charge of a particle is given by:q = ne

~F =1

4⇡✏0

|q1||q2|r2

r

|~r| =p(2L)2 + (2L)2

=p8L2

=p8L …(1)

~r has x- and y- components

~r = xi+ yj

Plug in their values:

~r = �2Li+ 2Lj

r2L

2L

We are also need r and r =~r

|~r|

…(2)= 2L(�i+ j)

Physics 1308: General Physics II - Professor Jodi Cooley

Now we can get r

r =~r

|~r|

|~r| =p8L

~r = 2L(�i+ j)

=2L(�i+ j)p

8L

Now, let’s put it all together.

~F =1

4⇡✏0

|q1||q2|r2

r = k(1e)(1e)

8L2[2p8(�i+ j)]

~F = (2.6⇥ 10�12 N)(�i+ j)

= (8.99⇥ 109 N ·m2/C2)(1.602⇥ 10�19 C)(1.602⇥ 10�19 C)

8(2⇥ 10�9 m)2[2p8(�i+ j)]

Physics 1308: General Physics II - Professor Jodi Cooley

The End (for Today!)

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