1. Electron Deflection How does a TV or computer monitor work? There are different technologies that can be employed, but one of the older ones is the use of a CRT (cathode ray tube). In a CRT, electrons are obtained by heating a wire, then accelerating these electrons up to a speed using a pair of charged plates, and then deflecting this electron beam by other charged plates.

2. Diagram AC screen +- V accelerating + - V deflecting L d h heater

3. Relations To find v x, use conservation of energy: ½ mv xo 2 + qV acc-o = ½ mv xf 2 + qV acc-f where v xo =0 and V acc-f – V acc-o = V accelerating so: (1/2)mv x 2 = q*V accelerating. If we know m, q, and V accelerating, we can determine v x. To find the height, h, we first realize that we have a constant electric field between the horizontal parallel plates that do the deflecting. If the Electric field is constant, so is force, and thus so is the acceleration (in the y direction).

4. Relations (1/2)mv x 2 = q*V accelerating, or v x = [2*q*V a /m] 1/2 V deflecting = E y * d F y = q * E y F y = m*a y, or a y = F y / m = q*E y /m = q*V deflecting / (m*d) = a y. v x = constant, so L = v x *t, or t = L/v x. a y = constant, so y = h = y o + v yo t+ (1/2)a y t 2, where y o = 0 and v yo = 0, so h = (1/2)a y t 2.

5. Review of relations v x = [2*q*V accelerating /m] 1/2 a y = q*V deflecting / (m*d) t = L/v x, or t 2 = L 2 /v x 2 = L 2 *m / [2*q*V accelerating ] h = (1/2)a y t 2 We can put all four relations together to form one relation connecting h, q, V deflecting, m, d, L, and V accelerating. In the computer homework, we are given: h, q, m, d, and L. We are asked for V deflecting and V accelerating. Since we now have one equation and two unknowns, we can choose one of them (say V accelerating, and then use the one equation to solve for V deflecting.

6. Result Since we have one (final) relation and two unknowns, we can then choose (within limits) one of the unknowns and then solve for the other. Since we are dealing with electrons, remember that electrons are attracted to the positive plate. This fact determines which plates should be positive and which plates should be negative.

7. Special case Please note that the equations we have developed will work for the case where the screen is right at the end of the deflecting plates. If the screen is placed a little further from the plates, we can determine the position of the beam at the screen by using the fact that the electron beam will go straight once it leaves the space between the deflecting plates.

8. Diagram AC screen +- V accelerating + - V deflecting L d h heater s yLyL

9. New case – theory Since we don’t know how high the object will be at the end of the plate like we did when the screen was right next to the deflecting plates, we can’t use exactly the same approach as we did for that case. The trick here is to recognize that for constant acceleration, y = ½ *a*t 2, and that a = (v f -v o )/t so that y = ½ *v f *t, or y = v f *(½*t). This means that the result of accelerating during the flight through the deflecting plates is the same as if the particle started at the center of the plate and kept a constant velocity equal to the final velocity when it was accelerating. The particle after leaving the plate does not accelerate anymore, so it does indeed travel along that same straight line.

10. New Case – theory continued If we start at the center of the deflecting plates and use that straight line, we then have to move a distance of h along the y while we move a distance of (½L+s) along the x (see the diagram). The proportion of distance in y versus x is constant since the speeds in v y and v x are both constant when we use that line, so y/x = y L /(½L) = h/(½L+s). We can now use the above relation to solve for y L. Now that we know y L, we can use the previous routine to solve for V defl once we choose V acc as long as we use y L in place of h.

11. Series and Parallel Practice with identification How are the elements (labeled E – could be either capacitors or resistors) connected below? How do we determine this? Are E 1 and E 2 connected in series in both? Are E 2 and E 3 connected in parallel in both? E1 E3E2 E1 E3

12. Series and Parallel Practice with identification In the circuit on the left, the current has to go through E 1 (E 1 is in series with whatever follows), but then the current has a choice of going through either E 2 or E 3 (so E 2 and E 3 are in parallel). E 1 is not in series with E 2, rather it is in series with the parallel combination of E 2 & E 3. E1 E3E2 E1 E3

13. Series and Parallel Practice with identification In the circuit on the left with the E’s being resistors, if R 1 = 8 Ω, R 2 = 2 Ω, and R 3 = 20 Ω, then the parallel combination R 2 & R 3 gives a resistance of 1/R 23 = [(1/2Ω) + (1/20Ω)] =.55/ Ω, so R 23 = 1.82 Ω, and the total resistance is R total = 8 Ω + 1.82 Ω = 9.82 Ω. E1 E3E2 E1 E3

14. Series and Parallel Practice with identification In the circuit on the right, the current has a choice to go through E 2 or E 3 (so parallel connection) but once it goes through E 2 it must go through E 1 (so E 1 and E 2 are in series). If it goes through E 3, it does not have to go through E 1. E 3 is in parallel with the series combination of E 2 and E 1. E1 E3E2 E1 E3

15. Series and Parallel Practice with identification In the circuit on the right, with the E’s being resistors, if R 1 = 8 Ω, R 2 = 2 Ω, and R 3 = 20 Ω, then the series combination R 1 & R 2 gives a resistance of R 12 = 8 Ω + 2 Ω = 10 Ω, and the total resistance is: 1/R total = [(1/10Ω) + (1/20Ω)] =.15/Ω so R total = 6.67 Ω. E1 E3E2 E1 E3