Find the electric potential at the origin due | vedclass

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(A) The electric potential $V$ at the origin due to a point charge $q$ at distance $x$ is given by $V = \frac{kq}{x}$.For the given distribution,the c

Vedclass · Accepted Answer

$kq \ln(2)$

Vedclass · Answer

(A) The electric potential $V$ at the origin due to a point charge $q$ at distance $x$ is given by $V = \frac{kq}{x}$.For the given distribution,the charges are $q$ at $x=1$,$-q$ at $x=2$,$q$ at $x=3$,$-q$ at $x=4$,and so on.The total potential at the origin is:$V = kq \left( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \right)$Using the Taylor series expansion $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$,for $x=1$,we get $\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$Therefore,$V = kq \ln(2)$.

Find the electric potential at the origin due to the given distribution of charges.

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