Electrostatics Electric Fields

Electric Field Strength Earth mass + proton Gravity Gravitational field found using a test mass FG = GmEm/r² mg = GmEm/r² g = GmE/r² Measured in N/kg Electrostatics Electric field found using a test proton FE = KQq/r² Eq = KQq/r² E = KQ/r² Measured in N/C

Electric Field Strength + Electric field is always found using a test proton Vector Measured in Newtons per Coulombs proton

Electric Field Strength Example What is the magnitude and direction of the electric field at the proton? E = KQ/r² E = (9x109Nm²/C²)(0.003C)/(1.5m)² E = +1.2x107N/C 3mC Positive means that test charge will experience repulsion + 1.5m

Electric Field Strength Example 2 What is the magnitude and direction of the electric field that the proton experiences? Strategy to Solve: Find Distances (radii) & Angles Find Electric Fields (E1 & E2) Vector Addition + 1m 4C -3C 1m 2m

Drawing Electric Fields A field is a collection of vectors showing an overall force at any point in space If you were to test the electric field with a proton in many places using the previous examples you would get a collection of vectors. + - proton proton proton Test particle is always a proton

Drawing Electric Fields Let’s test some harder situations: - + + + - -

Electric Field Strength 2 small point charges separated by a large distance Now, move the particles further apart: What happens to the electric force? FE = kQ1Q2/r² What happens to the electric field? E = kQ/r² Decreases Non-Uniform Electric Field Decreases + rf + ri

Electric Field Strength 2 large charged plates separated by a small distance Let’s use a test particle + + + + + - - - - + ri

Electric Field Between Parallel Plates Let’s look at a test particle: At Position 1: Whole lot of repulsion force from positive plate Little bit of attraction force from negative plate At Position 2: Average repulsion force from positive plate Average attraction force from negative plate At Position 3: Little bit of repulsion force from positive plate Whole lot of attraction force from negative plate Electric Force Is Uniform Everywhere FNET1 = FNET2 FNET1 = FNET2 = FNET3 FNET1 + + + + - - - - FEP2 FEP3 rP3 rN3 + 3 rP1 rN1 + 1 rP2 rN2 + 2 FEP1 FEN1 FEN2 FEN3

Electric Field Strength 2 large charged plates separated by a small distance Now, move the plates further apart: What happens to the electric force? What happens to the electric field? DO NOT USE FE and E EQUATIONS! Stays The Same Stays The Same Uniform Electric Field rf + + + + - - - - ri + + + + - - - -

Capacitor Electric Field Example Two parallel capacitor plates are 0.02m apart with an electric field of 600N/C between them. A proton is released from rest at point A, what force does the proton experience? What is its kinetic energy when it gets to point B? + + + + - - - - QP = +1.6x10-19C + A E = 600N/C B mP = 1.67x10-27kg 0.02m

Capacitor Electric Field Example What force does the proton experience? F = E * QP F = (600N/C)(1.6E-19C) F = 9.6x10-17N Find acceleration: a = F / m a = (9.6x10-17N) / (1.67x10-27kg) a = 5.75x1010m/s² + + + + - - - - 0.02m + A B QP = +1.6x10-19C mP = 1.67x10-27kg E = 600N/C

Capacitor Electric Field Example What is its kinetic energy when it gets to point B? Find final velocity: vf² = vi² + 2ad vf² = (0m/s)² + 2(5.75x1010m/s²)(0.02m) vf = 48000m/s Find kinetic energy: EK = ½mv² EK = ½(1.67x10-27kg)(48000m/s)² EK = 1.92x10-18J + + + + - - - - 0.02m + B A QP = +1.6x10-19C mP = 1.67x10-27kg E = 600N/C