Help With a Physics Question Involving Electrons and Gravitational Force

  • #1
Hi; Could someone please help me with this question: An electron (m= 9.11E-31 kg) is accelerated in the uniform field E (E= 1.45 x 10^4 N/C) b/w two parallel charged plates. The separation of the plates is 1.10cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate a)With what speed does it leave the hole? (I got the answer to this, which was 7.49E6 m/s. b) Show that the gravitational force can be ignored. I do not understand how to answer part b. Could somebody please help me. Thank you.

Answers and Replies

  • #2
Just show that the force of gravitation is so small that it can be ignored. Do you know the mass of the plate? If so,just use the force equation for gravity to find the net acceleration.

if not, you can just do a magnitudal analysis, the electron has mass of ~10^-31, the radius is ~10^-2 and G is 10^-11, using the equation for gravitational force, approximate the mass that the plate would need to accelerate the electron to that final speed. The mass should be somewhere in the range of 10^30kg (heavier than the mass of the earth).

You can then say that to accelerate the electron with 1/100th of the E field acceleration you would need a mass 100 times smaller,~ (10^28kg) which is still a couple thousand times heavier than earth.

I hope you can draw from this that gravity is much weaker than the EM force when it comes to the micro-scale.
  • #3
I think it's a bit simpler than that...

Haha, what is all this silliness about the mass of the plate or the radius of the electron? I think "gravitational force" simply means Earth's gravity, no?

We simply say, [tex] F_g = mg [/tex] and [tex] F_E = qE [/tex] .

Now, [tex] F_g \approx 10^{-29} N [/tex] and [tex] F_E \approx 10^{-15} N[/tex]. (The charge of the electron is about [tex]1.6 \times 10^{-19}C[/tex]).

Well, [tex] \frac{F_g}{F_E} \approx \frac{10^{-29}}{10^{-15}} = 10^{-14} [/tex].

This means the Earth's gravitational force is about 14 orders of magnitude weaker than the electric field of the plate, or one one hundredth trillionth. Negligible.
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  • #4
That makes sense too, and is probably the method they are lookin for. They really should be more specific though, as mine is just as correct given the vagueness of the problem :eek:.
  • #5
given the vagueness of the problem :eek:.

True... reminds me of what my professor says: grad students get to such advanced levels that when they come across a truly simple problem, they don't know how to deal with it because they've been doing the hard stuff for so long. :smile:

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