Electrostatics Electric Fields Electric Field Strength Earth mass

Electrostatics Electric Fields

Electric Field Strength Earth mass Gravity Gravitational field found using a test mass FG = Gm. Em/r² mg = Gm. Em/r² g = Gm. E/r² Measured in N/kg + proton 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 = (9 x 109 Nm²/C²)(0. 003 C)/(1. 5 m)² E = +1. 2 x 107 N/C Positive means that test charge will experience repulsion 3 m. C + 1. 5 m

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

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 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? Decreases FE = k. Q 1 Q 2/r² What happens to the electric field? Non-Uniform E = k. Q/r² Decreases Electric Field + + + rfi +

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 FNET 1 = FNET 2 = FNET 3 FEP 1 + + FEN 3 - 1 FEP 2 2 FEP 3 3 + + + FEN 1 FEN 2 r. P 1 r. P 2 r. P 3 r. N 1 r. N 2 r. N 3

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

Capacitor Electric Field Example Two parallel capacitor plates are 0. 02 m apart with an electric field of 600 N/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. 6 x 10 -19 C m. P = 1. 67 x 10 -27 kg + A + + E = 600 N/C 0. 02 m B -

Capacitor Electric Field Example What force does the proton experience? F = E * QP F = (600 N/C)(1. 6 E-19 C) F = 9. 6 x 10 -17 N Find acceleration: a=F/m a = (9. 6 x 10 -17 N) / (1. 67 x 10 -27 kg) + a = 5. 75 x 1010 m/s² A + + E = 600 N/C QP = +1. 6 x 10 -19 C B m. P = 1. 67 x 10 -27 kg 0. 02 m

Capacitor Electric Field Example What is its kinetic energy when it gets to point B? Find final velocity: vf² = vi² + 2 ad vf² = (0 m/s)² + 2(5. 75 x 1010 m/s²)(0. 02 m) vf = 48000 m/s Find kinetic energy: EK = ½mv² EK = ½(1. 67 x 10 -27 kg)(48000 m/s)² EK = 1. 92 x 10 -18 J + E = 600 N/C B + -19 C +1. 6 x 10 Q = + P + A m. P = 1. 67 x 10 -27 kg + 0. 02 m