A charge $+q$ fixed at each of the points $x=x_{0}$, $x=3 x_{0}, x=5 x_{0}, \ldots$ upto $\infty$ on $X$ -axis and charge $-q$ is fixed on each of the points $x=2 x_{0}, x=4 x_{0}$, upto $\infty$. Here $x_{0}$ is a positive constant. Take the potential at a point due to a charge $Q$ at $a$…

Question

A charge $+q$ fixed at each of the points $x=x_{0}$, $x=3 x_{0}, x=5 x_{0}, \ldots$ upto $\infty$ on $X$ -axis and charge $-q$ is fixed on each of the points $x=2 x_{0}, x=4 x_{0}$, upto $\infty$. Here $x_{0}$ is a positive constant. Take the potential at a point due to a charge $Q$ at $a$ distance $r$ form it to be $\frac{Q}{4 \pi \varepsilon_{0} r}$, then the potential at the origin due to above system of charges will be:

Aicharya · Accepted Answer

$\begin{aligned} V &=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{x_{0}}+\frac{q}{3 x_{0}}+\frac{q}{5 x_{0}}+\ldotsight] \+\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{(-q)}{2 x_{0}}+\frac{(-q)}{4 x_{0}}+\frac{(-q)}{6 x_{0}}+\ldotsight] \=& \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{x_{0}}\left[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\ldotsight] a \=& \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{x_{0}} \log _{e}(1+1)=\frac{q \log _{e} 2}{4 \pi \varepsilon_{0} x_{0}} \end{aligned}$ ^

A charge $+q$ fixed at each of the points $x=x_{0}$, $x=3 x_{0}, x=5 x_{0}, \ldots$ upto $\infty$ on $X$…

Solution