Electric Fields Review Coulomb’s Law Electric Fields Multiple-charge Electric Fields Continuous-charge Electric Fields Parallel Plate Fields Parallel Plate Examples Gauss’s Law

Coulomb’s Law Fundamental property charge Positive and negative charge Units Coulombs (C) Force Law

Compare with gravitation Fundamental property mass Positive mass only Units kilograms (kg) Force Law

“Field” concept (we did this in gravity) Newton’s Universal Law of Gravitation 𝐹= 𝐺 𝑚 1 𝑚 2 𝑟 2 One-step process (tedious) 𝐹 𝑥 = 𝐺𝑀 𝑚 𝑥 𝑟 2 𝐺𝑀 𝑚 𝑐𝑎𝑟 𝑟 2 𝐺𝑀 𝑚 𝑡𝑟𝑢𝑐𝑘 𝑟 2 𝐺𝑀 𝑚 𝑝𝑒𝑑 𝑟 2 Two-step process (simple) 𝑔= 𝐺𝑀 𝑟 2 =9.8 𝑁/𝑘𝑔 𝐹 𝑥 = 𝑚 𝑥 𝑔 𝑚 𝑐𝑎𝑟 𝑔 𝑚 𝑡𝑟𝑢𝑐𝑘 𝑔 𝑚 𝑝𝑒𝑑 𝑔

“Field” concept - Electrical Cellphone signal Multiple users sharing same tower. Why calculate each phone separately? 2 step process 1. Calculate common “field”. 2. Calculate each phone’s interaction with that “field”. “Field” equals # of “bars” you have!

Electrostatic vs. Gravitational Field In gravitation we calculate 𝐹= 𝐺𝑀 𝑟 2 𝑚 Bracket part becomes “g” In electrostatic we calculate 𝐹= 𝑘𝑄 𝑟 2 𝑞 Bracket part becomes “E” Field line point away from (+), toward (-) F=qE (+) moves with field, (-) moves against field

Electric Field Animation https://ef55311d-a-62cb3a1a-s-sites.googlegroups.com/site/physicsflash/Efield.swf?attachauth=ANoY7crCoduqPW1yrSvMtvv2qNVfA62NqyoNF1X8FY1ldipscty_-KXPmxmMyKYNdrruNd8vROoqOCxqee-i4LdS8Ct27vHfViZ597w8ETuqbnejUPkP7AiKqr4M-S3qn2VLzO2dQNl_KXv-Re0jQ5puGFnzKrPwZvuy0UkMwMU4QhXhJIIX4jWlWsQhQdgzqz-f17N3xQOXvEvfeWC1O-3ObHX89WV72w%3D%3D&attredirects=1

Electrostatic Field and Force Direction Define: Field of +q1 points outward, -q1 points inward Force on q2 (magnitude & direction) given by 𝐹= 𝑞 2 𝐸 Result: +q2 feels force with field, -q2 feels force against field So like charges repel, opposite charges attract + -

What if gravity was attractive/repulsive??? Positive and negative mass (!!) (don’t worry, there isn’t such a thing) “Negative” mass falls to ceiling “Positive” mass falls to floor g

Electrostatic vs. Gravitational Field Electric Field Gravitational Field Field Concept Can be anything Usually 9.8 m/s2 Force & field F=q2E Force in field/opposite direction F= m2g Force in field direction Field Definition Ratio of Force/charge Ratio of Force/mass, (simplifies to acceleration) Strength Electrostatic so strong appears on circuit-board scale Gravity so weak only appears on planetary scale Comment “E” quite interesting Varies all over the place “g” usually boring usually constant 9.8 m/s2 Superposition Can superimpose continuously Wires, electrodes, circuit boards Can superimpose discretely Planets, etc

Electric Field - Example 16-8 (1) Field from Q1 𝐸 1 =𝑘 𝑞 1 𝑟 1 2 =9∙ 10 9 𝑁 𝑚 2 𝐶 2 25∙ 10 −6 𝐶 0.02 𝑚 2 =5.625 ∙ 10 8 𝑁/𝐶 𝑡𝑜 𝑙𝑒𝑓𝑡 Field from Q2 𝐸 2 =𝑘 𝑞 2 𝑟 2 2 =9∙ 10 9 𝑁 𝑚 2 𝐶 2 50∙ 10 −6 𝐶 0.08 𝑚 2 =0. 703125∙ 10 8 𝑁 𝐶 𝑡𝑜 𝑙𝑒𝑓𝑡 𝐸 𝑡𝑜𝑡𝑎𝑙 =6.3 ∙ 10 8 𝑁 𝐶 𝑡𝑜 𝑙𝑒𝑓𝑡

Electric Field - Example 16-8 (2) Force on proton at P 𝐹=𝑞𝐸= +1.6∙ 10 −19 𝐶 6.3 ∙ 10 8 𝑁 𝐶 =1∙ 10 −10 𝑁 𝑡𝑜 𝑙𝑒𝑓𝑡 Acceleration of proton at P 𝑎= 𝐹 𝑚 = 1∙ 10 −10 𝑁 1.67∙ 10 −27 𝑘𝑔 =5.99 ∙ 10 16 𝑚 𝑠 𝑡𝑜 𝑙𝑒𝑓𝑡 Force on electron at P 𝐹=𝑞𝐸= −1.6∙ 10 −19 𝐶 6.3 ∙ 10 8 𝑁 𝐶 =1∙ 10 −10 𝑁 𝑡𝑜 𝑟𝑖𝑔ℎ𝑡 Acceleration of electron at P 𝑎= 1∙ 10 −10 𝑁 9.11∙ 10 −31 𝑘𝑔 =1.09 ∙ 10 20 𝑚 𝑠 𝑡𝑜 𝑟𝑖𝑔ℎ𝑡

Electric Field - Example 16-9 (1) Get magnitudes 𝐸 2 =𝑘 𝑄 2 𝑟 2 2 = 9∙ 10 9 𝑁 𝑚 2 𝐶 2 50∙ 10 −6 𝐶 0.3 𝑚 2 =5∙ 10 6 𝑁/𝐶 𝐸 1 =𝑘 𝑄 1 𝑟 1 2 = 9∙ 10 9 𝑁 𝑚 2 𝐶 2 50∙ 10 −6 𝐶 0.6 𝑚 2 =1.25 ∙ 10 6 𝑁/𝐶

Electric Field - Example 16-9 (2) XY table Field X-component Y-component E2 0 N/C +5 x 106 N/C E1 +1.1 x 106 N/C -0.625 x 106 N/C Total +4.375 x 106 N/C

Continuous charge distributions Coulomb’s Law plus a lot of integral calculus! (don’t worry, I’ll just show results) Field of line of charge Field of ring of charge Field of sheet of charge Field between 2 sheets of charge

1- Field of Line of Charge

2 – Field of Ring of Charge

3 – Field of Sheet of Charge

4- Continuous charge distributions - Summary

5 – Field between 2 Sheets of Charge In region between parallel plates 𝐸= 𝑄 𝜀 𝑜 𝐴 = 𝜎 𝜀 𝑜 𝑄=𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒 𝜎=𝑐ℎ𝑎𝑟𝑔𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒 𝐴=𝐴𝑟𝑒𝑎 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑝𝑙𝑎𝑡𝑒 𝜀 𝑜 = 1 4𝜋𝑘 = 8.85∙ 10 −12 𝐶 2 𝑁 𝑚 2 Note: now the field is constant Coulomb’s Law + Lots of integral calculus = Constant fields 𝐸= 𝑘𝑄 𝑟 2 𝑑𝑥 𝜎 𝜀 𝑜 (and we just show results)

Field between 2 Sheets of Charge In region between parallel plates 𝐸= 𝑄 𝜀 𝑜 𝐴 = 𝜎 𝜀 𝑜 Note field is constant between plates Uses for parallel plates Capacitors (charge storage) Beam accelerators / deflectors (old TVs, high-energy labs) Semiconductor junctions Photocopy machines

Example - Photocopy Field Electrostatic for twice weight 𝑞𝐸−2𝑚𝑔=0 𝐸= 2𝑚𝑔 𝑞 = 2 9∙ 10 −16 𝑘𝑔 9.8 𝑚 𝑠 2 20 1.6∙ 10 −19 𝐶 =5500 𝑁/𝐶

Examples Problem 23 Problem 24 𝐹=𝑞𝐸= −1.6∙ 10 −19 𝐶 2360 𝑁 𝐶 =−4.16∙ 10 −16 𝑁 𝑤𝑒𝑠𝑡 Problem 24 𝐸= 𝐹 𝑞 = 3.75∙ 10 −14 𝑁 1.6∙ 10 −19 𝐶 =234,000 𝑁 𝐶 𝑠𝑜𝑢𝑡ℎ

Examples 𝐹=𝑚𝑎 Problem 27 Problem 31 𝐹=𝑞𝐸= −1.6∙ 10 −19 𝐶 750 𝑁 𝐶 =−1.2∙ 10 −16 𝑁 𝑎= 𝐹 𝑚 = −1.2∙ 10 −16 𝑁 9.1∙ 10 −31 𝑘𝑔 =1.32∙ 10 14 𝑚 𝑠 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 Problem 31 𝐹=𝑚𝑎 𝐸= 𝐹 𝑞

Field Lines Field lines start at (+) charge, end at (-) charge. Without charge, field lines don’t exist. A (+) test charge moves in the direction of the field lines, a (-) test charge moves opposite the direction of field lines. The cross-sectional density of field lines is proportional to the electric field strength. 𝐸∝ 𝐿𝑖𝑛𝑒𝑠 𝐴𝑟𝑒𝑎

Gauss’s Law Number of field lines entering or leaving any volume proportional to charge enclosed. 𝐸∙𝑑𝐴 𝑐𝑜𝑠𝜃= 𝑞 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 𝑜 Evaluate over entire closed surface Field lines perpendicular to surface Need to know symmetry

Gauss’s Law Examples Point charge Line of charge Sheet of charge

Gauss’s Law 1 – Point charge 𝐸∙𝑑𝐴 𝑐𝑜𝑠𝜃= 𝑞 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 𝑜 For spherical surface 𝐸 4𝜋 𝑟 2 = 𝑞 𝜀 𝑜 𝐸 = 𝑞 4𝜋 𝑟 2 𝜀 𝑜

Gauss’s Law 2 – Line charge 𝐸∙𝑑𝐴 𝑐𝑜𝑠𝜃= 𝑞 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 𝑜 For cylindrical surface 𝐸 2𝜋𝑟𝑙= 𝜆𝑙 𝜀 𝑜 𝐸 = 𝜆 2𝜋𝑟𝜀 𝑜

Gauss’s Law 3 – Sheet charge 𝐸∙𝑑𝐴 𝑐𝑜𝑠𝜃= 𝑞 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 𝑜 For pillbox surface 𝐸 2𝐴= 𝜎𝐴 𝜀 𝑜 𝐸 = 𝜎 2𝜀 𝑜