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Saurabh_Shimpi · Jan 12, 2011, 03:05 AM

protons and electron are placed at point x=1cm, 3cm 5cm & x=2cm,4cm,6cm respectivelyon x-axis up to infinity what is potential difference origin

jcaron2 · Jan 12, 2011, 08:21 AM

We know the potential from each charge is of the form

$V=\frac{k\cdot Q}{r}$

Assuming charges of +q for the protons and -q for the electrons, you'll get contributions from the protons as follows:

$V_{protons}=\frac{k\cdot q}{1}+\frac{k\cdot q}{3}+\frac{k\cdot q}{5}+...$

and contributions from the electrons as follows:

$V_{electrons}=-\frac{k\cdot q}{2}-\frac{k\cdot q}{4}+\frac{k\cdot q}{6}+...$

Summing all the contributions together, we get

$V=\frac{k\cdot q}{1}-\frac{k\cdot q}{2}+\frac{k\cdot q}{3}-\frac{k\cdot q}{4}+... = k\cdot q\sum _{n=1}^{\infty}\frac{\left (-1 \right )^{n+1}}{n}=k\cdot q\cdot \ln \left (2 \right )$

That last jump where the summation was replaced by ln(2) comes from the known value of the alternating harmonic series. See this link for a more in-depth explanation.

jcaron2 · Jan 12, 2011, 01:09 PM

By the way, since the charge locations are specified in cm, make sure your factor k is scaled to those units, not meters.