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Question about electric potential energy - need confirmation

  • Thread starter Kelvin
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This is a homework problem of a EM course:

Calculate the potential energy, per ion, for an infinite one-dimensional ionic crystal, that is, a row of equally spaced charges of magnitude e and alternating sign.
Hint: The power series expansion of ln(1+x) may be useful.


Here's my calculation:

Let [tex]u_n[/tex] be the potential energy when there is n charges.
[tex]u_1 = 0[/tex]

[tex]u_2 = \frac{1}{4 \pi \epsilon_0} \frac{-e^2}{d}[/tex] where d is the separation of two charges

[tex]u_3 = \frac{1}{4 \pi \epsilon_0} \left( 2\times \frac{-e^2}{d} + \frac{e^2}{2d}\right)[/tex]

[tex]u_4 = \frac{1}{4 \pi \epsilon_0} \left( 3 \times \frac{-e^2}{d} + 2 \times \frac{e^2}{2d} + \frac{-e^2}{3d}\right)[/tex]

....

[tex]u_n = \frac{1}{4 \pi \epsilon_0} \left( (n-1) \times \frac{-e^2}{d} + (n-2) \times \frac{e^2}{2d} + \dots + \frac{(-1)^{n-1} e^2}{(n-1)d}\right)[/tex]

[tex] = \frac{e^2}{4 \pi \epsilon_0 d} \sum_{i=1}^{n-1} \frac{(-1)^i (n-i)}{i}[/tex]

[tex] = \frac{e^2}{4 \pi \epsilon_0 d} \sum_{i=1}^{n-1} (-1)^{i-1} \left( 1- \frac{n}{i} \right) [/tex]

[tex] = \frac{e^2}{4 \pi \epsilon_0 d} \left( \sum_{i=1}^{n-1} (-1)^{i-1} - n \sum_{i=1}^{n-1}\frac{(-1)^{i-1}}{i} \right)[/tex]

potential energy per ion
= [tex]u_n/n = \frac{e^2}{4 \pi \epsilon_0 d} \left( \frac{1}{n}\sum_{i=1}^{n-1} (-1)^{i-1} - \sum_{i=1}^{n-1}\frac{(-1)^{i-1}}{i} \right)[/tex]

as n tends to infinity,

[tex]u_n/n = \frac{e^2}{4 \pi \epsilon_0 d}\left( 0 - \ln 2 \right)[/tex]

I am not sure my last 2 steps are correct or not. The series [tex]\sum_{i=1}^{n-1}(-1)^{i-1}[/tex] diverges as n tends to infinity. If we divide the series by n, do we get really zero?

Thanks for your help!
 

Answers and Replies

  • #2
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While you probably don't need this since it was 6 years ago: I just completed this problem on a practice GRE test and the solution is easier if you read the question closely. It asks for the potential energy per ion, as opposed to the entire system. This allows the summation to fall out with a ln(2) included. Hope this helps to future readers!
 

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