Expert: Paul Klarreich - 2/6/2007

Question
hi there mr. klarreich, i've been stuck on this problem for quite some time, nearly 3 days now and i have not made much progress, probably because i'm not too sure where to start; please help!

the question is:
Electrons repel each other with a force which is inversely proportional to the square of the distance between them; call the proportionality constant 'k' in the units to be used. Suppose one electron is fixed at x=0 on the x-axis.

a. find the work done in moving a second electron along the x-axis from the point x = 10 to the point x = 1.
b. find the work done in moving the second electron along the x-axis from the point x = M to the point x = 1.

thank you very much for the help :)

Answer
Questioner:   Aliza
Category:  Calculus
 
Subject:  Integral Calculus on Work
Question:  hi there mr. klarreich, i've been stuck on this problem for quite some time, nearly 3 days now and i have not made much progress, probably because i'm not too sure where to start; please help!

the question is:
Electrons repel each other with a force which is inversely proportional to the square of the distance between them; call the proportionality constant 'k' in the units to be used. Suppose one electron is fixed at x=0 on the x-axis.

a. find the work done in moving a second electron along the x-axis from the point x = 10 to the point x = 1.
b. find the work done in moving the second electron along the x-axis from the point x = M to the point x = 1.

thank you very much for the help :)
...........................................
Hi, Aliza,

As I recall, Work = Force * distance.  In this case,

F(x) = k/x^2

a. If you move from  x = 10 to x = 1, the work should be:

{x=1   k dx
|      ---- =
}x=10  x^2

{x=1   
|   k x^-2 dx =
}x=10   

k x^-1   - k
------ = ----,  from  x = 10  to  x = 1
- 1       x

= -k[1/1 - 1/10]

= -k[9/10] = -9k/10

Now I suppose the work comes out negative because you are doing work ON the electron; I leave the interpretive part to you.

b. For moving from  x = M  to x = 1, it's the same, but you do:

k x^-1   - k
------ = ----,  from  x = M  to  x = 1
- 1       x

= -k[1/1 - 1/M]
    M - 1
= -k[-----]
      M

[We really should do part b first.  Then just put M = 10 and get the answer to a.]