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Homework Help: Curve STRAIGHTENING ARGH!

  1. Nov 29, 2004 #1

    Yes, curve straightening....every high school student's favorite thing.... :confused:

    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:

    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)?
  2. jcsd
  3. Nov 29, 2004 #2


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    Gold Member

    Well, in a word: yes.

    You've already answered your own question haven't you???

  4. Nov 29, 2004 #3
    I was just wondering if I did everything right....

    That's all....cause it seemed complicated at first....
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