The Motion of Charged Particle in Electric Fields Problem

AI Thread Summary
The discussion centers on calculating the smallest possible distance between two electrons fired at each other at high speed. Participants express concern over discrepancies between their calculations and the book's solution, particularly regarding the formula used. There is confusion about the inclusion of a square in the distance variable 'r' in the energy equation. The correct approach involves using the energy equation, which is energy = qV = Fr = kq1q2/r, rather than the force equation. Overall, the conversation highlights potential errors in the textbook's methodology while affirming its overall value.
Freeq
Messages
7
Reaction score
0
3. Two electrons are fired at 3.5x10^6 m/s directly at each
other.
(a) Calculate the smallest possible distance between the
two electrons.

(b) Is it likely that two electrons in this situation will actually
get this close to each other if the experiment is
performed? Explain your answer.

Hello, let's look at question 3(a). I just did that question, but I got different answer than the one in the book.
There is the book solution is in the attachment.

I'm concerned about a mistake in the formula. I think that there is no square in r. Even two pages after, you can see that there is no square there. How is this Possible?
 

Attachments

  • dsa2.jpg
    dsa2.jpg
    10.6 KB · Views: 592
  • dsa3.jpg
    dsa3.jpg
    19.5 KB · Views: 530
Physics news on Phys.org
Hi Freeq! :smile:
Freeq said:
I'm concerned about a mistake in the formula. I think that there is no square in r. Even two pages after, you can see that there is no square there. How is this Possible?

Yes, you're right. :smile:

The dimensions there are completely wrong (and the units at the end combine to m1/2, not m). :frown:

energy = qV = Fr = kq1q2/r not kq1q2/r2 :rolleyes:

(which book is this?)​
 
Nelson 12 Physics. It looks like they used formula for electric force, but not energy. However, I still think it is a good book.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top