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U7 l9-motion ofcharge


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  • 1. Unit 7 – Lesson 9 Motion of a Charge Due to Electric Forces The Cathode Ray Tube Defining the Electron VoltWorkbook Reference Pages: 38 - 39
  • 2. In this lesson, we will look at how a charged particle can be made to move in 3 dimensions by using three electric fields. This is the technology used in a cathode ray tube (old style TV- the heavy box).Acceleration of a Charge and the q/m RatioThe first step is to get the charge moving, to do this an accelerating potential difference (ΔV) is used to generate a uniform electric field. The diagram, on the next slide, shows how this would be drawn.
  • 3. Accelerating a Charge from RestIn the diagram, a charge is at rest at “A” but is accelerated by the electric field to “B” where it Δ V leaves the field.F = qE, But a = (q/m)EThe magnitude of the electric field intensity (E) is found from: A BE = ΔV/Δd, -q DOMSo, a = (q/m)(ΔV/Δd), the importance of the q/m ratio on Δ d the magnitude of “a” should be apparent.
  • 4.  Recall that a + charge will move in the direction of the electric field arrows and a – charge will move in the opposite direction. To move a charge (q) from side to side and up and down, two additional electric fields must be added to deflect the charge in these directions. This is shown in the diagram for a cathode ray tube (CRT) setup. H o r iz o n ta l S c re e n V e r tic a l F ie ld F ie ld Top Far S id eUP -q DO M Δ d C e n tr e L in e Near B o tto m S id e Δd H E le c t r o n G u n (fo r a C R T ) P u r p le a r r o w s in d ic a te D O M o f d e fle c tio n
  • 5. Understanding the CRT (see diagram from previous slide) The e- gun, generates and accelerates the e- (from rest) to its final horizontal speed (vH), which can be calculated by considering conservation of _________. Or, ½mv² = qΔV Rearrange to get: vH = √(2(q/m)ΔV){This speed is less than c (speed of light) and ranges from approximately 2.5 x 106 m/s to 9.0 x 107 m/s } The horizontal plates generate a vertical electric field, so the e- will be deflected up or down. The field direction can be quickly changed if required. Since the charge is accelerated only up or down, vH will _____ change and only a vertical velocity component will be generated (vV)
  • 6. Calculating vVSince vH does not change, we can calculate the amount of time it take to pass through the horizontal plates Δt = ( ΔdH / vH ), refer to the diagram on slide # 4.Calculate a⊥ using the equation on slide # 3, and using the information for the horizontal plates ( vertical field). Then vV = a⊥ Δt , where a⊥ = (q/m)(ΔV/Δd) = (q/m)(|E|)Note: We will not perform any calculations for the vertical plates (the 3rd dimension of motion). All problems will be 2-D.
  • 7. After emerging from the horizontal plates, the velocity of the e- would be vH + vV , using vector addition (Pythagorean thm would be required as well as finding the direction of the e- ).Defining the Electron Volt (eV) This is a measure of energy If an electron is accelerated from rest through a potential difference of 1.0 Volt its energy would be equal to 1.0 eV: 1.0 eV = q ΔV = 1.6 x 10-19 C x 1.0 V 1.0 eV = 1.6 x 10-19 J, {this is the Joule equivalent}
  • 8. Example:a. The accelerating potential difference for an e- gun is 2.0 x 104 V. Determine the eastward speed of the e- when it emerges from the gun. [8.4 x 107 m/s]b. If the e- then enters an upward electric field of magnitude 5.0 x 104 N/C calculate a ⊥. Is it likely Fg has any meaningful effect on the motion of the e- ? [8.8 x 1015 m/s2 [Down], no]c. If the horizontal length of the plate is 9.0 cm, determine the final velocity of the e- when it emerges from the horizontal plates. [8.4 x 107 m/s [E 60 Down]]
  • 9. Practice Questions Textbook problems: Page 359 #21 – 22 Page 376 # 55, 61, 75Answer for 75 is incorrect, should be 58.4 mm Workbook Problems: Page 37 # 1- 4, 7, 10