Ionic Drive Problem

I am having trouble with my calculated values for part B of the problem. In part A I used the formula F=qE, where F is the force, q is the charge = 1.6 * 10^-19 coulombs and E is the electric field = 10,000 V/m. From that formula I calculated F = 1.6 * 10^-15 N.

On part B I used F=ma (rearranged into a = F/m), using the given mass and the force calculated in part A to get an acceleration of a = 7.272727273 * 10^7 m/s^2. I then plugged the acceleration and given length into the kinematic equation Δx = 0.5at^2 (ignoring v_0t because I assumed it starts from rest), giving me a time of t = 7.41619849 * 10^-5 seconds. I then plugged that into the Impulse formula, ΔP = FΔt, using the initial calculated force and I got, Impulse = 1.19197315 * 10^-19. So, my question is, is my calculated impulse value too small? I do not have any frame of reference in the area of Ionic drives to know if that is even in the ballpark of being correct or acceptable.

  • $\begingroup$ +1 This is a reasonable ion propulsion question. Even though it's an exercise or homework-like you've shown all work and as far as I can tell gotten the correct answer. $\endgroup$
    – uhoh
    May 5 at 0:35
  • $\begingroup$ This site, and Stack Exchange in general, is not for answering homework questions. If there is a fundamental aspect that you don't understand about the topic, please submit another question with a more specific title. $\endgroup$
    – Freddie R
    May 5 at 0:36
  • 1
    $\begingroup$ @FreddieR I apologize. I wasn't really trying to get someone to give me an answer to the actual problem. I was more so just trying to understand, from a conceptual point of view, if my value was reasonable for an Impulse and to do so I thought it would be best to show the specific situation it came from. I will refrain from posting homework problems in the future. $\endgroup$
    – AstroNOT98
    May 5 at 0:40
  • $\begingroup$ @uhoh Also, thank you very much for the feedback. $\endgroup$
    – AstroNOT98
    May 5 at 0:41
  • $\begingroup$ @FreddieR the OP did not ask for an answer to a homework question. Have you read through the question carefully enough to see that yet? The OP posted their answer and included all work and described all assumptions, and asks only for confirmation that these can be applied to ion engines. $\endgroup$
    – uhoh
    May 5 at 0:50

tl;dr: Yes you've got it all right as far as I can tell.

An atomic mass unit is about 1.66E-27 kg so the mass discussed there (2.2E-23 kg) is about 13,000 AMU which is quite a hefty molecule or a nanoparticle with many hundred atoms.

The force is given by

$$F = qE$$

With $q$ in coulombs and $E$ in volts/meter the force will be in Newtons and indeed it will be 1.6E-15.

With a mass $m$ the acceleration $F/m$ will indeed be 7.27E+07 m/s^2.

Starting with the familiar $x = \frac{1}{2} a t^2$ and $v = a t$ we can use

$$t = \sqrt{\frac{2x}{a}}$$

to get 7.41E-05 seconds, and 5.39 km/s final velocity.

A different way to get the same answer

Impulse from one particle is also the momentum of that particle (since momentum is conserved) so $mv$ is 1.187E-19 kg m/s, within round-off error of your 1.19197315 * 10^-19 kg m/s.

  • 1
    $\begingroup$ If you know the atomic mass unit with three digits, there is no need to calculate the time with four digits and the speed with five digits. $\endgroup$
    – Uwe
    Jun 5 at 8:16
  • $\begingroup$ @Uwe I see what you mean, please feel free to edit and improve the answer! $\endgroup$
    – uhoh
    Jun 5 at 10:55

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