Time it takes for an electron to travel a set distance

[Reprisal35]

Homework Statement

In a CRT tube, electrons are accelerated by a 20,000 V potential difference between the electron gun (the cathode) and the positive metal mesh 5.00 cm away.
a. What is the electron’s speed when it reaches the positive wire mesh?
b. How much time does it take the electron to reach the wire mesh?

Homework Equations

$U = q∆V\\ K = \frac{1}{2}mv^2\\ F = ma\\ F = qE \\ E = \frac{V}{d}\\ d = d_0 + (v_0)(t) + \frac{1}{2}(a)(t^2)\\$

The Attempt at a Solution

For part A, I got the following which is actually the same example we did in class.

$U = K\\ q∆V = \frac{1}{2}mv^2\\ (1.6*10^{-19}C)(20000\frac{J}{C}) = \frac{1}{2}(9.1*10^{-31}kg)(v^{2})\\ 3.2*10^{-15}J = 4.55*10^{-31}kg(v^{2}) \\ (3.2*10^{-15}J)/(4.55*10^{-31}kg) = v^{2} \\ 7.033*10^{15}\frac{m^{2}}{s^{2}} = v^{2} \\ v = √7.033*10^{15}\frac{m^{2}}{s^{2}} \\ v = 8.386*10^{7}\frac{m}{s}\\$

Where I'm getting a bit paranoid is part B.

At first I thought, in order to get $t$, all we needed to do was use the formula, $t = \frac{d}{v}$ and I got an answer of $5.96*10^{-10}s$, but that seems to simple for this kind of problem.

So then I decided to use a combination of several formulas we learned in the past in order to determine the time it takes for an electron to travel a set distance.

Here's what I did:

$F = ma\\ F = qE \\ E = \frac{V}{d}\\ ma = \frac{qV}{d}\\ a= \frac{qV}{dm}\\ d = d_0 + (v_0)(t) + \frac{1}{2}(a)(t^2)\\ d = \frac{1}{2}(a)(t^2)\\ d = \frac{1}{2}[\frac{qV}{dm}](t^2)\\ 2d = \frac{qV}{dm}(t^2)\\ \frac{2d^{2}m}{qV} = t^2\\ t = √\frac{2d^2m}{qV}\\ t = √\frac{(2)(.05m)^{2}(9.1*10^{-31}kg)}{(1.6*10^{-19}C)(20,000\frac{J}{C})}\\ t = √\frac{4.55*10^{-33}}{3.2*10^{-15}}\\ t = √1.422*10^{-18}s\\ t = 1.19*10^{-9}s\\$

Now looking back at my second attempt, it seems a bit ridiculous to do all that to get $t$. So my question is, is my second attempt for getting $t$ correct or did I mess up?

 

[PeterO]

Homework Statement

In a CRT tube, electrons are accelerated by a 20,000 V potential difference between the electron gun (the cathode) and the positive metal mesh 5.00 cm away.
a. What is the electron’s speed when it reaches the positive wire mesh?
b. How much time does it take the electron to reach the wire mesh?

Homework Equations

$U = q∆V\\ K = \frac{1}{2}mv^2\\ F = ma\\ F = qE \\ E = \frac{V}{d}\\ d = d_0 + (v_0)(t) + \frac{1}{2}(a)(t^2)\\$

The Attempt at a Solution

For part A, I got the following which is actually the same example we did in class.

$U = K\\ q∆V = \frac{1}{2}mv^2\\ (1.6*10^{-19}C)(20000\frac{J}{C}) = \frac{1}{2}(9.1*10^{-31}kg)(v^{2})\\ 3.2*10^{-15}J = 4.55*10^{-31}kg(v^{2}) \\ (3.2*10^{-15}J)/(4.55*10^{-31}kg) = v^{2} \\ 7.033*10^{15}\frac{m^{2}}{s^{2}} = v^{2} \\ v = √7.033*10^{15}\frac{m^{2}}{s^{2}} \\ v = 8.386*10^{7}\frac{m}{s}\\$

Where I'm getting a bit paranoid is part B.

At first I thought, in order to get $t$, all we needed to do was use the formula, $t = \frac{d}{v}$ and I got an answer of $5.96*10^{-10}s$, but that seems to simple for this kind of problem.

So then I decided to use a combination of several formulas we learned in the past in order to determine the time it takes for an electron to travel a set distance.

Here's what I did:

$F = ma\\ F = qE \\ E = \frac{V}{d}\\ ma = \frac{qV}{d}\\ a= \frac{qV}{dm}\\ d = d_0 + (v_0)(t) + \frac{1}{2}(a)(t^2)\\ d = \frac{1}{2}(a)(t^2)\\ d = \frac{1}{2}[\frac{qV}{dm}](t^2)\\ 2d = \frac{qV}{dm}(t^2)\\ \frac{2d^{2}m}{qV} = t^2\\ t = √\frac{2d^2m}{qV}\\ t = √\frac{(2)(.05m)^{2}(9.1*10^{-31}kg)}{(1.6*10^{-19}C)(20,000\frac{J}{C})}\\ t = √\frac{4.55*10^{-33}}{3.2*10^{-15}}\\ t = √1.422*10^{-18}s\\ t = 1.19*10^{-9}s\\$

Now looking back at my second attempt, it seems a bit ridiculous to do all that to get $t$. So my question is, is my second attempt for getting $t$ correct or did I mess up?

Can you show your working to get your answer in the bit I made red, so we can check you did it correctly.

 

[Reprisal35]

Can you show your working to get your answer in the bit I made red, so we can check you did it correctly.

Sure,

$d = 0.05m\\ v = 8.386∗10^7\frac{m}{s}\\ v = \frac{d}{t}\\ t = \frac{d}{v}\\ t = \frac{0.05m}{8.386∗10^7\frac{m}{s}}\\ t = 5.96*10^{-10}s$

 

[PeterO]

Sure,

$d = 0.05m\\ v = 8.386∗10^7\frac{m}{s}\\ v = \frac{d}{t}\\ t = \frac{d}{v}\\ t = \frac{0.05m}{8.386∗10^7\frac{m}{s}}\\ t = 5.96*10^{-10}s$

OK, You have used the final velocity, not the average velocity.

 

[Reprisal35]

OK, You have used the final velocity, not the average velocity.

Well that makes sense now.

$v_{avg} = \frac{v_0 + v_f}{2}\\ v_{avg} = \frac{0 + 8.386∗10^7}{2}\\ v_{avg} = 4.193*10^7\\ t = \frac{0.05}{4.193*10^7}\\ t = 1.192*10^{-9}s$

I feel kinda silly for forgetting to take the avg of speed... Thanks for pointing that out!

So both methods give the same answer. Is it safe to assume both methods work and that the answer is correct?

 

[PeterO]

Well that makes sense now.

$v_{avg} = \frac{v_0 + v_f}{2}\\ v_{avg} = \frac{0 + 8.386∗10^7}{2}\\ v_{avg} = 4.193*10^7\\ t = \frac{0.05}{4.193*10^7}\\ t = 1.192*10^{-9}s$

I feel kinda silly for forgetting to take the avg of speed... Thanks for pointing that out!

So both methods give the same answer. Is it safe to assume both methods work and that the answer is correct?

Seems reasonable to me.

 

[Reprisal35]

Seems reasonable to me.

Awesome! Thanks again!