Motion of a Charged Particle in an Electric Field Physics 12

Joke of the day:

Review:

1. If you have a charge of 1.0C, how many electrons have been removed from the object? 2. What voltage is created if a charged particle is placed 2.5cm from a -1.0mC charge? 3. What electric field is created between two plates, separated by 5.0cm with a voltage of 250V? 6.25x V 5 000N/C

Answers: x V N/C

Electron in a uniform electric field: 1. Draw a uniform electric field (contains a positive and negative plate) where the field is directed into the ground 2. Draw a free body diagram for an electron placed in a uniform electric field where the field is directed into the ground FeFe FgFg

Consider: What happens when a charged particle is placed (with zero initial velocity) inside a uniform electric field? Draw a picture to show an electron placed at the negative plate that is creating a uniform field FeFe

What if we wanted to find the velocity of the electron? A derivation…..using formulas you know! V=Ed, E =F onq, F=ma, Vf 2 = V o 2 +2ad q Reminder: m e = 9.1x kg, q e = 1.6x C

Example #1: If an field is created using 250V with plates separated by 1.0cm, determine how fast the electron is moving when it gets to the positive plate (ignore gravity)?

Problem #1: You’ve been hired by Stanford to determine the voltage required to accelerate an electron or positron to 0.95c a. What voltage is required if the accelerator is 3.2km long? b. Find the electric field required in order to accomplish this final velocity for one of the particles. Answers: 2.3x10 5 V, 72 N/C

Crooke’s Tube: Invented by Sir William Crookes (1875) consisting essentially of a sealed glass tube from which nearly all the air has been removed and through the walls of which are passed two electrodes. When a high voltage is applied between the two electrodes, electrons are emitted from the cathode and are accelerated toward the anode. Many of these electrons (aka cathode rays), miss the anode and strike instead the glass wall of the tube, causing it to exhibit fluorescence.

Review: Loss of potential energy is equal to a gain in kinetic energy -ΔPe = Δke Electric potential (V) is a type of potential energy! We can use this to calculate the velocity of particles…. Ke = ½ mv 2

Problem #2: a. How fast would an electron be moving in Crooke’s tube if: V = 5000V d = 0.025m b. How much kinetic energy would the electrons have? Answers: 4.2x10 7 m/s, 8.0 x J

LHC Interested?? Large Hadron Collider Zj27IM Zj27IM

Problem #3: The first stage of the LHC accelerates protons to.75c determine: a. The voltage required if the electric field exists over 1.5km b. The electric field required Answers: 1.4x10 5 V, 96 N/C

To do: In class Assignment