Welcome to Physics 1308 The Electric Field and Motion Charles-Augustin de Coulomb 1736-1806
Announcements Assignments for Tuesday, September 4th: - Reading: Chapter 23.1-23.3 & 23.5-23.6 - Watch 2 Videos: https://youtu.be/iwu1mhdmmv4 Lecture 6 - Gauss Law AND Video: https://youtu.be/3yb0z5edcby Lecture Vectors Part 2 - Homework 3 Assigned - due before class on Tuesday, September 11th. Dr. Cooley will be out of the office Friday, September 7. Her office hours are canceled for that day.
Review Question 1 Four charges are located on the corners of a square as shown in the drawing. What is the direction of the net electric field at the point labeled P? A) toward the upper left corner of the square B) toward the middle of the right side of the square C) toward the middle of the bottom side of the square D) toward the lower right corner of the square E) There is no direction. The electric field at P is zero.
Key Concepts If a particle with charge q is placed in an external electric field E, an electrostatic force F acts on the particle: ~F = q ~ E The torque on an electric dipole of dipole moment p when placed in an external electric field E is given by a cross product: ~ = ~p ~ E A potential energy U is associated with the orientation of the dipole moment in the field, as given by a dot product: U = ~p ~E
Review of PHYS 1307/1303 Equations of Motion v = v 0 + at x = x 0 + v 0 t + 1 2 at2 v 2 = v 2 0 +2a(x x 0 ) x = x 0 + v v 0 2 t
Question 1 Consider the drawing, where the solid lines with arrows represent the electric field due to the charged object. An electron is placed at the point P and released from rest. Which of the following vectors represents the direction of the force, if any, on the electron?
Question 2 A positively charged object is located to the left of a negatively charged object as shown. Electric field lines are shown connecting the two objects. The five points on the electric field lines are labeled A, B, C, D, and E. At which one of these points would a test charge experience the largest force? Answer: D ~F = q ~ E E / 1 r 2
Instructor Problem: E-ink Devices
E-INK (Electrophoretic Ink
https://en.wikipedia.org/wiki/e_ink
Instructor Problem: Turn the Page Titanium dioxide particles coated with polymer to give them a net charge: q = 16e m =6.7 10 15 kg Time taken to turn the page - that is, move particles from the bottom to the top (or vice versa) of the cell is t = t t 0 =0.25 s What electric field strength is required to achieve this?
First sketch the problem and take account of what was given and what you need to find: Given: q = 16e Δy E m =6.7 10 15 kg t = t t 0 =0.25 s y = 40.0µm = 40.0 6 m q Find the E needed to achieve this Δt and make the page turn in an e-ink based device. We know that the magnitude of the E-field is related to the force through the following equation. ~ F = q ~ E From Newton s Second Law we can relate force and acceleration. Thus, we can relate a to E. q E ~ = m ~a E ~ = m ~a q ~ F = m ~a
We know it takes time to move the titanium dioxide up to the top of the cell. Use EOM to relate Δt, Δy, Plug into equation for E : ~ E = m ~a q a. ~ E = m q apple 2 y ( t) 2 y = y 0 + v 0 t + 1 2 ~a t2 y y 0 = v 0 t + 1 2 ~a t2 Finally, plug in the numbers: v0= 0 m/s, thus ~ E = 6.7 10 15 kg 16 1.6 10 19 C apple 2(40.0 10 5 6 m) (0.25 s) 2 y = 1 2 ~a t2 ~ E =3.4 N/C ~a = 2 t 2 y This corresponds to voltages of about 0.1 mv, perfect for low-power, long-battery life devices!
Student Problem: DNA Gel Electrophoresis DNA Electrophoresis is the process of using a uniform electric field in a medium to separate the various genes in a strand of Deoxyribonucleic Acid, or DNA. The sugar-phosphate backbone of the nucleic acid chain is negatively charged, and can be accelerated with an electric field. Fragments move at different rates due to their sizes and the porosity of the medium in which they have to move (a gel).
DNA Gel Electrophoresis A certain gene has an electric charge q = -3.4 x 10-14 C and a mass m = 6.0 x 10-20 kg. The gene must traverse an agarose gel whose porosity results in a drag force of 2.6 x 10-27 N. a) If this gene begins at rest and is subjected to an external uniform electric field whose strength is E = 1.0 10-13 N/C, how long (in minutes) does it take the gene to move 1.0 cm? b) What is the kinetic energy of the gene by the time it reaches the 1.0 cm mark?
First sketch the problem and take account of what was given and what you need to find: Given: ~ E =1.0 10 13 N/C Fdrag E ~ F drag =2.6 10 27 N d = 1 cm q = 3.4 10 14 C gene m =6.0 10 20 kg Need to find how long it takes to move the gene 1 cm. The gene is initially at rest, but we know it is accelerated by the Force if feels by the electric field. However, that force is opposed by the drag force of the gel. The total force is thus, ~F total = ~ F field + ~ F drag ~ F total = ~ F field ~ F drag and the field force is given by ~ F field = q ~ E = 3.4 10 14 C 1.0 10 13 N/C =3.4 10 27 N Substitute into Ftotal ~ F total =8.0 10 28 N
The acceleration can be written as ~a = ~ F total m = 8.0 10 28 N 6.0 10 20 kg =1.3 10 8 m/s 2 Now use the EOM. We are given the distance traveled and the initial velocity. x = x 0 + v 0 t + 1 2 at2 x x 0 = v 0 t + 1 2 at2 d = 1 2 at2 t = r 2d a = s 2(0.010 m) 1.3 10 8 m/s 2 = 1224.7 s Convert to minutes. 1224.7 s 1 min 60 sec! t = 20 m
b) What is the KE at d = 1.0 cm? KE = 1 2 mv2 need the speed v We can get speed from the EOM. v 2 = v 2 0 +2a(x x 0 ) v 2 =2ad Substitute into definition of KE KE = 1 2 m(2ad) = mad =(6.0 10 20 kg)(1.3 10 8 m/s 2 )(0.01 m) KE =8.0 10 30 J
The End for Today! FoxTrot