How can I linearize the equation for q/m in terms of V?

[GodsChild086]

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)?

 

[cepheid]

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)?

Well, in a word: yes.

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

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'

 

[GodsChild086]

I was just wondering if I did everything right...

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