1.

There is an infinite straight chain of alternating charges `q` and `-q` . The distance between teh neighbouring charges is equal to `a`. Find the interaction energy of each charge with all the others. Instruction . Make use of the expansion of In `(1 + alpha)` in a power series in `alpha`.

Answer» As the chain is of infinite length any two charge of same sign will occur symmetrically to any other charge of opposite sign.
So, interaction energy of each chagre with all the others,
`U = -2 (q^(2))/(4pi epsilon_(0) a) [1 - (1)/(2) + (1)/(3) - (1)/(4) + .....` up to `oo`] ....(1)
But In `(1 + x) = x - (1)/(2) x^(2) + (1)/(3) x^(3)`..... up to `oo`
and putting `x = 1` we get IN `2 = 1 - (1)/(2) + (1)/(3) + ....` up to `oo` .......(2)
From Eqs (1) and (2),
`U = (-q^(2) In 2)/(4pi epsilon_(0) a)`


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