Question

There is a infinite Staight chain of alternate charges q and -q .the distance between the two neighbouring charges is equal to e find the interaction energy any charge with all the other charges​

Answer 1

Question :

There is a infinite Staight chain of alternate charges q and -q .The distance between the two neighbouring charges is equal to a. find the interaction energy any charge with all the other charges.

Theory :

• Electric potential energy due to two point charges :

${\purple{\boxed{\large{\bold{</p><p>U= \frac{k \: q_{1} \: q_{2}}{r} }}}}}$

• electric potential energy due to system of point charges

${\red{\boxed{\large{\bold{U = U _{12} + U _{13} + U _{13} + U _{14}.......}}}}}$

Sepcial series :

${\purple{\boxed{\large{\bold{log(1 + x) = x - \frac{x {}^{2} }{2} + \frac{x {}^{3} }{3} - \frac{x {}^{4} }{4} + \frac{x {}^{5} }{5} + ... \infty }}}}}$

Solution :

$U= U_{12} + U _{13} + U_{13} + U _{14}....... \infty$

$\implies \: u = \frac{k \times q \times - q}{a} + \frac{k \times q \times q}{2a} + \frac{k \times \: q \times - q}{3a} + \frac{k \times q \times q}{4a} + .... + \infty$

$\implies \: u = \frac{ - kq {}^{2} }{a} (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \infty )$

$\implies \: u = - \frac{kq {}^{2} }{a} log_{e}(2)$

${\purple{\boxed{\large{\bold{U = - \frac{kq {}^{2} }{a} log_{e}(2) }}}}}$

it is the required solution!