Welcome Back to Physics Electric Fields. Micheal Faraday Physics 1308: General Physics II - Professor Jodi Cooley

Transcription

1 Welcome Back to Physics 1308 Electric Fields Micheal Faraday

2 Announcements Assignments for Thursday, August 30th: - Reading: Chapter Watch Video: Lecture 4 - Electric Fields of Charge Distributions Homework 2 Assigned - due before class on Tuesday, September 4th. Dr. Cooley will be out of the office the week of August 27th. Her office hours are cancelled for this week. Please her if you have any immediate concerns or needs. Mr. James Thomas is lecturing in her place.

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4 Review Question 1 Three insulating balls are hung from a wooden rod using thread. The three balls are then individually charged via induction. Subsequently, balls A and B are observed to attract each other, while ball C is repelled by ball B. Which, if any, of the following statement(s) concerning this situation is correct? A) A and B are charged with charges of opposite signs; and C is charged with charge that has the same sign as B. B) A and B are charged with charges of the same sign; and C is electrically neutral. C) A is electrically neutral; and C is charged with charge that has the same sign as B. D) B is electrically neutral; and C is charged with charge that has the same sign as A.

5 Key Concepts Electric Field: The electric field E at any point is defined in terms of the electrostatic force F that would be exerted on a positive test charge q0 placed there: Since we know ~F = 1 q 1 q r 2 ˆr We can write he magnitude of the electric field E set up by a particle with charge q at distance r from the particle as

6 Key Concepts Electric Field Lines: What happens if I bring 2 positive charges close to each other?

7 Question 1 The drawing shows a hollow conducting sphere with a net positive charge uniformly distributed over its surface. A small negatively-charged object has been brought near the sphere as shown. What is the direction of the electric field at the center of the sphere? A. There is no electric field at the center of the sphere B. To the left C. To the right D. Upward E. Downward

8 Question 2 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 N/C.

9 Problem 1: Repelling Charges What is the strength of the electric field at point P (green dot)? The horizontal and vertical scales of the grid are 1.0 nm per grid step. The two charges are the same: q1 = q2 = -2.5 x 10-6 C.

10 First sketch the problem: P Given: q 1 = q 2 = ~r 1 = (13.0 nm)î + (10.0 nm)ĵ ~r 2 =( 2.0 nm)î +(5.0 nm)ĵ r1 We need to find the net electric field nm ~E 1 = k q 1 r 2 1 ˆr 1 ~ E2 = k q 2 and r 2 2 ˆr 2 Calculate r1 13 nm r 1 = p nm = p 269 nm 2 ˆr 1 = ~r r 1 = 13 p 269 î + 10 p 269 ĵ r2 5 nm Calculate r2 ˆr 1 =(0.792)î +(0.610)ĵ P 2 nm r 2 = p ( 2) 2 +( 5) 2 nm = p 29 nm ˆr 2 = ~r r 2 = 2 p 29 î + 5 p 29 ĵ

11 ˆr 1 =(0.792)î +(0.610)ĵ Now add them together: ˆr 2 =( 0.371)î +( 0.928)ĵ ~E = ~ E 1 + ~ E 2 = kq[ ˆr 1 r 2 + ˆr 2 r 2 ] = kq[( )î +( )ĵ] =( N m 2 /C 2 )( C)[( )î +( )ĵ] ~E =[2.3 î +6.7 ĵ] N/C Check the math!!!

12 Problem 2: Target Field What is the magnitude of q required to achieve an electric field at point P with magnitude E = 3.0 x 10-6 N/C? ~E neg = k q Need r, r 2, and ˆr r 2 neg ˆr neg ~r neg = L î + L ĵ = L( î + ĵ) r neg = D = p 2L ˆr = ~r r = L(î + ĵ) p 2L = 1 p 2 (î + ĵ) Plugging into Eneg, ~E neg = kq 2L 2 1 p2 (î + ĵ) = kq 2 p 2L 2 ( î ĵ) (1) D = 40.0 nm L = D p 2

13 Now for Epos: ~E pos = k q r 2 pos ˆr pos Need r, r 2, and ˆr ~r pos = L î + L ĵ = L( î + ĵ) r pos = D = p 2L ˆr = ~r r = L( î + ĵ) p 2L = 1 p 2 ( î + ĵ) Plugging into Epos, ~E pos = p kq 2 2L ( î + ĵ) Combine the fields: ~E = E ~ neg + E ~ pos = p kq 2 2L ( î ĵ)+ kq 2 p 2L ( î + ĵ) = kq p 2L ( î) q = p 2LE k But the question asked for magnitude, so we drop the minus sign and solve for q. q = C

14 The End for Today! State of New South Wales, Department of Education and Training, 2008