Curve STRAIGHTENING....ARGH! Yes, curve straightening....every high school student's favorite thing.... So in my Physics class, our teacher did a demo with the CRT (cathode ray tube) and then he assigned us with the lab exercise. Here are the components of what I'm supposed to include: Here's the equation presented when Thomson discovered the charge to mass ratio of an electron: v²/Br = q/m ; where: v = velocity (m/s) B = magnetic field strength (T) q = charge on electron (C) m = mass (kg) r = radius (m) Challenge: Find the equation for q/m in terms of V. Okay, this part I got. Equations I used: Ep=Ek qV = 1/2mv² and the other equation used: v²/Br = q/m Using the first question I solved for v²: qV = 1/2mv² ; v² = 2qV/m My next step taken: V/Br = q/m ; q²/m² = v²/B²r² q²/m² = 2qV/mB²r² Here is the equation for q/m in terms of V: q/m = 2V/B²r² Then we played around with the cathode ray tube, and told us to make a chart. It's for each time he increases the diameter of the circle (beam of electrons), we record the voltage. This is another challenge he set for us: Write the equation so that q/m is the slope. Here's what I did....I'm not sure if I linearized it right (bad grammar....English is not my first language, so I apologize for this). So I started with this equation again: q/m = 2V/B²r² Equation for a straight line is: y = mx + b In this case, I don't have a 'b', so it's just y = mx My equation (to linearize) I came up with is: y = mx 2V=(q/m)(B²r²) ; where: 2V is the 'y part' (q/m) is the 'm part' (slope) and (B²r²) is the 'x part' After coming up with our equation, he said that make a new table of values. With our new table of values, he said to make a straight-line graph. So my question is (after going through all this mess of typing), is my table going to have a column for 2V and B²r²? As for the graph, is it going to be 2V vs. B²r²? (Where 2V is on the y-axis and B²r² is on the x axis)?