Question

There is an infinite straight chain of alternating charges $q$ and $-q$. The distance between the two neighbouring charges is equal to $a$. Find the interaction energy of any charge with all the other charges.

• A. $\frac{2q^2}{4\pi {\epsilon }_{0}a}$
• B. $\frac{2q^{2}lo{g}_{e}2}{4\pi {\epsilon }_{0}a}$
• C. $-\frac{2q^{2}lo{g}_{e}2}{4\pi {\epsilon }_{0}a}$
• D. none of these

Answer 1

The correct option is C $-\frac{2q^{2}lo{g}_{e}2}{4\pi {\epsilon }_{0}a}$

Here the the interaction energy of first charge (q) with all the other charges is

$U=U_{12}+U_{13}+U_{14}+U_{15}+......$

or $U=\frac{q(-q)}{4\pi {ϵ}_{0}a}+\frac{q\left(q\right)}{4\pi {ϵ}_{0}2a}+\frac{q(-q)}{4\pi {ϵ}_{0}3a}+\frac{q\left(q\right)}{4\pi {ϵ}_{0}4a}+.....$

$=\frac{-q^2}{4\pi {ϵ}_{0}a}[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....]$

$=\frac{-q^2}{4\pi {ϵ}_{0}a}lo{g}_{e}2$ as $lo{g}_e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$ or $lo{g}_e(1+1)=lo{g}_{e}2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....$

The interaction energy of charge at middle of chain with any charge will be $2U$ because this time both end of chain extends to infinite.