Electric Fields of Charge Distributions

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1 Welcome to Physics 308 Electric Fields of Charge Distributions Charles-Augustin de Coulomb Physics 308: General Physics II - Professor Jodi Cooley

2 Announcements Assignments for Tuesday, September 4th: - Reading: Chapter Watch Video: ecture 5 - The Electric Field and Motion 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. Physics 308: General Physics II - Professor Jodi Cooley

3 Review Question Consider the electric field lines shown in the drawing. Which of the following statements correctly describes this situation? A. The electric field is due to a positively charged particle. B. The electric field is due to a negatively charged particle. C. The electric field is due to a positively charged particle and a negatively charged particle. D. The electric field is due to particles that are both charged either positively or negatively. Physics 308: General Physics II - Professor Jodi Cooley

4 Key Concepts Electric Dipole: An electric dipole consists of two particles with charges of equal magnitude q but opposite signs, separated by a small distance d. The magnitude of the electric field set up by an electric dipole at a distant point is given by Physics 308: General Physics II - Professor Jodi Cooley

5 Electric Field due to a ine of Charge Charge distribution: If the charge is distributed uniformly over the whole length then we can write the charge density per unit length as = Q The charge is uniform over the entire length. Thus, = Q = q and if Δq > dq (an infinitesimal bit of charge and Δ > d (an infinitesimal bit of length = dq d Physics 308: General Physics II - Professor Jodi Cooley

6 Summing the fields due to infinitesimal charges: Think of the line as being composed of a large number of tiny pieces of Q, each of equal size dq. When we add them up, they equal Q. Q = dq + dq + dq + dq +...dq Z Q = 0 dq Each piece, dq, emits a piec of the total electric field. So we can write the field emitted by dq as de. Then Z ~ Etotal ~E total = 0 d ~ E Each piece is a field due to a point charge, dq. d ~ E = k dq r 2 ˆr Physics 308: General Physics II - Professor Jodi Cooley

7 Now, write the field due to one dq as drawn above: We need r, r 2 and ˆr in terms of this coordinate system. ~r = î + D ĵ r = p 2 + D 2 ˆr = î + D ĵ p 2 + D 2 Substitute into the point charge electric field equation: d ~ E = k dq r 2 ˆr = k dq ( 2 + D 2 î + D ĵ p 2 + D 2 = k dq ( 2 + D 2 [ î + D ĵ] Physics 308: General Physics II - Professor Jodi Cooley

8 Recall: d ~ E = k dq ( 2 + D 2 [ î + D ĵ] Substitute = dq d d ~ E = k dq = d d ( 2 + D 2 [ î + D ĵ] Now, sum up the pieces to get the total - and y- components. ~E = Z = k î d ~ E = Z + Z + k î( ( 2 + D 2 d 2/3 ( ( 2 + D 2 d 2/3 Write the - and y- components separately. look this integral up in a book to find: d ~ E =[k ( 2 + D 2 î]d Z + ( 2 + D 2 2/3 = ( 2 + D 2 /2 + d ~ E y =[k D ( 2 + D 2 ĵ]d = p + p =0 Thus, ~E =0î Physics 308: General Physics II - Professor Jodi Cooley

9 Similarly for Ey: ~E y = k Dĵ Z + ( 2 + D 2 d 2/3 Thus, ~E y = k Dĵ D 2 apple + p 2 + D 2 look this integral up in a book to find: = k Dĵ D 2 apple ( Z + ( 2 + D 2 2/3 d = D 2 ( 2 + D 2 /2 + ~E y = 2k D ĵ Finally, put it together: Case:! + p 2 + D 2! p 2 = = ~E = ~ E + ~ E y =0î + 2k D ĵ The field lines radiate Case:! outward from the line. Strength falls off linearly p 2 + D 2! p 2 = = with distance D. ~ E line = 2k D Physics 308: General Physics II - Professor Jodi Cooley

10 Problem Find the electric field at point P as illustrated in the diagram below. Point P is located a distance D above a line of uniformly distributed charge with length. ayout the key steps: - choose a representative point charge, dq, and write its electric field de - make de a function of coordinates and geometry - integrate de and solve for E. Uniform line of charge means that Q = dq d = This is constant! Physics 308: General Physics II - Professor Jodi Cooley

11 Choose a differential element of a charge: d ~ E = k dq r 2 ˆr ( Need dq, r 2, ˆr in terms of geometry and coordinate system. ~r = î + y ĵ! r = p 2 + D 2 =( 2 + D 2 /2! ˆr = î + D ĵ ( 2 + D 2 Since = Q = dq d! dq = d = Q d Now, substitute into ( d ~ E = kq î + D ĵ ( 2 + D 2 d Physics 308: General Physics II - Professor Jodi Cooley

12 Integrate: ~E = Z d ~ E = Z 0 apple kq î + D ĵ ( 2 + D 2 d = apple Z 0 kq apple Z 0 ( 2 + D 2 d î+ kq D ( 2 + D 2 d ĵ Eamine -component: ~E = kq î Z 0 ( 2 + D 2 d ook up in a book of integrals ~E = kq î ( 2 + D 2 /2 0 = kq î apple D + ( 2 + D 2 /2 = kq apple î ( 2 + D 2 /2 + D D( 2 + D 2 /2 Physics 308: General Physics II - Professor Jodi Cooley

13 Eamine y-component: ~E = kq î apple ( 2 + D 2 /2 + D D( 2 + D 2 /2 ~E y = kqd ĵ Z 0 ( 2 + D 2 d ook up in a book of integrals ~E y = kqd ĵ apple D 2 ( 2 + D 2 /2 0 = kqd ĵ D 2 apple 0+ ( 2 + D 2 /2 = kq ĵ D( 2 + D 2 /2 = kqĵ D( 2 + D 2 /2 Now, put it together. ~E = ~ E + ~ E y ~E = kq D( 2 + D 2 /2 apple D ( 2 + D 2 /2 î + ĵ Physics 308: General Physics II - Professor Jodi Cooley

14 The End for Today! State of New South Wales, Department of Education and Training, 2008 Physics 308: General Physics II - Professor Jodi Cooley

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

Welcome Back to Physics Electric Fields. Micheal Faraday Physics 1308: General Physics II - Professor Jodi Cooley Welcome Back to Physics 1308 Electric Fields Micheal Faraday 1791-1867 Announcements Assignments for Thursday, August 30th: - Reading: Chapter 22.3-22.5 - Watch Video: https://youtu.be/wc79wv5klx4 Lecture