1 / 5

Today 2/21

Today 2/21. E-Field and Coulomb’s Law 18.4-5 Examples HW: “2/21 2-D E Field ” Due Wednesday 2/26. q 1. k Q s. k Q s. k 6 q o. k 2 q o. 1. 1. kq o. kq o. r 2. r 2. (4 r o ) 2. (6 r o ) 2. 6. 8. r o 2. r o 2. 4. kq o 2. 24. r o 2. 1. 1. kq o. 1. kq o. 6. 8. r o 2.

hunter-stark
Download Presentation

Today 2/21

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Today 2/21 • E-Field and Coulomb’s Law 18.4-5 • Examples • HW: “2/21 2-D E Field ” Due Wednesday 2/26

  2. q1 k Qs k Qs k 6qo k 2qo 1 1 kqo kqo r2 r2 (4ro)2 (6ro)2 6 8 ro2 ro2 4 kqo2 24 ro2 1 1 kqo 1 kqo 6 8 ro2 24 ro2 E = kQs /r2 Many Charge Example (Like HW) Each Square is ro on a side, Q1 = -4qo, Q2 = +2qo, Q3 = -6qo (See how this simplifies the math) Q1 Find the E field at Q1’s location. Up, (away from Q2) EQ2,x = Q2 EQ3,x = Down, (toward Q3) Q3 FOthers,q1 = q1Enet,x Add vectors to get Enet,x Down at the x

  3. q2 k Qs k Qs k 6qo k 4qo 3 1 kqo kqo r2 r2 (4ro)2 (2ro)2 4 2 ro2 ro2 5 kqo2 2 ro2 3 1 kqo 5 kqo 2 4 ro2 4 ro2 E = kQs /r2 Many Charge Example (Like HW) Each Square is ro on a side, Q1 = -4qo, Q2 = +2qo, Q3 = -6qo (See how this simplifies the math) Q1 Find the E field at Q2’s location. Up, (toward Q1) EQ1,x = Q2 EQ3,x = Down, (toward Q3) Q3 FOthers,q2 = q2Enet,x Add vectors to get Enet,x Down at the x

  4. = k(2qo)/(2ro)2 E = kQs /r2 Many Charge Example What is the force on +4qo? (direction also) +qo +4qo Find the field at +4qo due to the “other” charges. These are the “source” charges. ro E+q = kQs/r2 = k(qo)/(ro)2 = Eo -2qo -3qo Direction? Away from + so right E-2q = kQs/r2 = k2qo/2(ro)2 = k(qo)/(ro)2 = Eo E-3q = kQs/r2 = k(3qo)/(ro)2 = 3kqo/(ro)2 = 3Eo

  5. Etotal = 3.72 + 0.292 Eo E = kQs /r2 Many Charge Example What is the force on +4qo? (direction also) +qo +4qo E+q = Eo E-2q = Eo E-3q = 3Eo ro Now add E-Field vectors -2qo Ex = E+q,x + E-2q,x -3qo Ex = Eo - Eosin45 Ey = E-2q,y + E-3q,y Ex = 0.29Eo Ey = Eosin45 + 3Eo  = tan-1(0.29/3.7) = 4.5° Ey = 3.7Eo Fon +4q = (4qo)3.71Eo = 14.84qoEo = 3.71Eo Fon +4q = 14.84kqo2/ro2 same direction as Etotal

More Related