Continuous Charge Distributions Physics 212 Lecture 2 Today's Concept: The Electric Field Continuous Charge Distributions

Coulomb’s Law (from last time) 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 q1 q2 q3 q4 q1 q2 q3 q4 F2,1 F3,1 F4,1 F1 Add E = infront of F1/q1 +q1 -> -q1  direction reversed MATH: 33

Electric Field The electric field E at a point in space is simply the force per unit charge at that point. Electric field due to a point charged particle Superposition Perhaps picture of glowing point charge It is called a “field” because it is a continuous function defined at every point in space. q2 E4 E2 E Field points toward negative and Away from positive charges. E3 q4 q3 08

Checkpoint “I don't completely understand how to determine the direction of the electric field.” Score at A is lower.. Don’t vote but use simulatoin to demonstrate what is going on B? 4) 09

Checkpoint E Draw the fields 12

Two Charges Both charges are negative Both charges are positive Two charges q1 and q2 are fixed at points (-a,0) and (a,0) as shown. Together they produce an electric field at point (0,d) which is directed along the negative y-axis. y (0,d) This should be moved forward…. but it is on another page in the book  q1 q2 (-a,0) (a,0) x Which of the following statements is true: Both charges are negative Both charges are positive The charges are opposite There is not enough information to tell how the charges are related 22

- - + + + - 23

Checkpoint A B C D Force on q from +Q: F(from +Q) = + k Qq/r2 (Acts to the right.) Force on q from +2Q: F(from +2Q) = -k 2Qq/(2r)2 (Acts to the left.) Total force on q : + k Qq/r2 -k 2Qq/(2r)2 = + k Qq/2r2 So it acts to the right. 12

Example d +q -q P What is the direction of the electric field at point P, the unoccupied corner of the square? Need to know d Need to know d & q (A) (B) (C) (D) (E) Calculate E at point P. 20

Continuous Charge Distributions Summation becomes an integral of vectors. Integrate over all infinitesimal charges dq. r is vector from dq to the point P where we want to find E. P Line of charge: charges l = Q/L dE r dq = ldx 25

Charge Density What has more net charge?. Some Geometry Linear (l=Q/L) Coulombs/meter Surface (s = Q/A) Coulombs/meter2 Volume (r = Q/ V) Coulombs/meter3 What has more net charge?. A sphere w/ radius 2 meters and volume charge density r = 2 C/m3 A sphere w/ radius 2 meters and surface charge density s = 2 C/m2 Both A) and B) have the same net charge. 28

Checkpoint 10) EA = 0 EB  0 29

Calculation What is ? (A) (B) (C) (D) (E) y P x y a h P r dq=ldx Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is l C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? We know: What is ? (A) (B) (C) (D) (E) 33

Calculation What is ? (A) (B) (C) (D) q2 P y x y a h P r dq=ldx q1 q2 q2 Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is l C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? We know: What is ? (A) (B) (C) (D) 33

Calculation cosq2 DEPENDS ON x !! What is ? (A) (B) (C) y a h P r dq=ldx q1 q2 q2 Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is l C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? We know: What is ? (A) (B) (C) none of the above cosq2 DEPENDS ON x !! 33

Calculation What is ? (A) (B) (C) (D) q2 P y x y a h P r dq=ldx q1 q2 q2 Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is l C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? We know: What is ? (A) (B) (C) (D) 33

Calculation What is ? q2 P y x y a h P r dq=ldx q1 q2 q2 Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is l C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? We know: What is ? 33

Observation x y a h P r dq=ldx q1 q2 q2 Charge is uniformly distributed along the x-axis from the origin to x = a. The charge denisty is l C/m. What is the x-componen t of the electric field at point P: (x,y) = (a,h)? Note that our result can be rewritten more simply in terms of q1. Exercise for student: Change variables: write x in terms of q Result: obtain simple integral in q 33

Prelecture 3 and Homework 1 due by 8:00 AM tomorrow! Labs start Thursday, June 14.