1. The problem statement, all variables and given/known data Question: A cathode ray tube in a TV has an electron gun operating at 10 kV. a: What is the velocity (in m.s-1) at which the electrons are emitted? Answer:5.9x107 b: How many volts are needed between the deflector plates to move the beam impact spot across a 70 cm screen if the screen is 120 cm from the deflector? 2. Relevant equations Tan(θ)=vE/vI 1/2mv2 QE=qV 3. The attempt at a solution First attempt: So first I worked out the angle the beam has to move which was 30.3 degrees. Then I used the equation Tan(angle) = velocity in the direction of the field/ initial velocity. Which was Tan(30.3) X 5.9x107 = 3.4x107 Then I used 1/2 x 9.1x10-31 x (3.4x107 )2 to get 5.3x10-16 <I think I'm going wrong here. When I read through the lecture slides things become limited. then PE = qV, so PE/q = V, so 5.3x10-16 / 1.6x10-19 = 3287 Then I was given this advice: " The problem is that velocity is a vector quantity, while energy is not. So it seems that you appropriately worked out the velocity in the direction of the field. (I'll call this ve, and the initial velocity vi) What you should do then is find the total velocity vtot=sqrt( ve2 + vi 2 ). You then use this number to find the total KE of the electron. Subtract the initial KE (find with vi) and this gives you the energy (PE) which must be added by the electric field." So I tried this, sqrt(5.9x1072+3.4x1072) = 6.81x107 So Total KE = 0.5x9.1x10-31x6.81x1072 =2.11x10-15 Initial KE = 0.5x9.1x10-31x5,9x1072 = 1.583855x10^-15 2.11x10-15 - 1.583855x10-15 = 5.26145x10-16 Then QE=qV, so QE/q = V So 5.26145x10-16/1.6x10-19 = 3288.40625 But this is also wrong.