 How should an electric field be position so that it will exert an equal and opposite electric force on the moving electron? Estimate the size of the field needed to do this for the values given in question 6..
 Thomson realized that with the correct orientation of the electric and magnetic fields he could measure the velocity of the electrons without knowing either their mass or charge. When the electric and magnetic field are arranged in the correct orientation and their values adjusted so that the net force on the electron is 0 (no deflection) then the condition is true.
 In the applet the magnetic field is controlled by varying the current in the coil that surrounds the tube while the electric field is controlled by varying the voltage potential across the two plates. The applet will calculate the respective magnetic and electric fields produced by the current and voltage potential. You can record various combinations of current and voltage potential that result in a zero force or nodeflection condition on the electrons. To do this press the "record data" button located on the lower right corner of the applet. You can retrieve your data by pressing the "options" button (upper menu bar) and then press "data". Find 3 different currentvoltage potential combinations that result in zero deflection of the beam. Use these measurements to complete the following table and estimate the speed of the electrons in the simulation in the horizontal or "xdirection".
Magnetic Field B (T) 
Electric Field E (V/m) 
velocity v_{x} (m/s) 










A better way to determine the velocity of the electrons is to graph E as a function of B. Collect at least 5 sets of data similar to those in question 9 and use this to complete a graph similar to the one shown below. What shape of graph do you expect and what feature of the graph will tell you the velocity of the electrons? (Note  you can copy the data from the applet and paste it directly into EXCEL for further analysis and graphing). What value did you find for the velocity of the electrons?
Part Three  Determining the ChargetoMass Ratio
Knowing the speed of the electrons is crucial! Now that you know this, it is relatively straight forward to determine the ratio between the charge on the electron and its mass. In this section it is assumed that you already know the speed of the electrons. We no longer need the magnetic field and hence assume B = 0 T for the following.
 Why can we "ignore gravity" in this experiment? The simple answer is that eventhough the electrons are pulled downward by their interaction with the Earth's gravitational field, the actual time for this interaction is much too short for any appreciable change in the vertical momentum of the electron. To show that this is so, determine the following:
 The time the electrons will take to travel across the cathode ray tube. Use the speed that you determined in part 2 and also estimate the distance that they travel (length of the catode ray tube) to be 0.50 m.
 What downward velocity does an object acquire if it falls for the amount of time that you calculated in part a?
 Explain, in your own words why these calculations suppourt the claim that gravitational effects can be ignored in this experiment.
 Now that you know you can ignore the effects of gravity, study the following diagram and explain why the electron's velocity is changing in the way that is shown. This shows the electron moving between the charged plates. Why is the velocity changing in the ydirection only?
 Show that the time required for the electrons to travel across the plates ( a distance 'd') is and that by the time the electron leaves the plates it has acquired a downward velocity given by the expression .
 What force causes the electron to deflect downward? (Remember  you concluded earlier that it's not gravity!).
 Show that as the electron leaves the region between the parallel plates, its velocity in the downward direction is given by the formula .
 You're almost there! Now show how the angle of deflection of the beam relates the velocity components in the x and y directions. Draw simple vector diagram representing the x and y velocities and show that .
 This is the "key" that relates the angle of deflection to the chargetomass ratio . Show that the equations in questions 15 and 16 can be combined to give the following expression: .
Part Four  Recreating the ChargetoMass data
You are now ready to simulate one of the great experiments of the early 20th century! In collecting the data be sure to set B = 0 mT. Collect enough data to enable good averaging of the results (minimum of 5 data sets). Tabulate your data as shown below. (It would be a god idea to do this in EXCEL)
In this simulation we are using d = 0.055 m
Trial # 
E (V/m) 
Angle of Deflection 
q/m (C/kg) 
1 



2 



3 



4 



5 



average 

