An infinite number of charges each equal to $4 \mu \mathrm{C}$ are placed along $X$-axis at $x=1 m ;

Question

An infinite number of charges each equal to $4 \mu \mathrm{C}$ are placed along $X$-axis at $x=1 m ; x=2 m$; $x=4 m ; x=8 m$ and so on. The field at the origin due to all of them is\

Professor PC Thomas Classes · Answer

Field $=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}}=9 	imes 10^{9} \frac{\mathrm{q}}{\mathrm{r}^{2}} \quad($ in SI $)$ $$ \begin{aligned} 	herefore \mathrm{E} & =\mathrm{E}_{1}+\mathrm{E}_{2}+\mathrm{E}_{2}+\ldots . \ & =9 	imes 10^{9}\left[\frac{\mathrm{q}}{1^{2}}+\frac{\mathrm{q}}{2^{2}}+\frac{\mathrm{q}}{4^{2}}+\ldotsight] \ & =9 	imes 10^{9} \mathrm{q}\left[1+\frac{1}{4}+\frac{1}{16}+\ldotsight] \ & =9 	imes 10^{9} 	imes\left(4 	imes 10^{-6}ight)\left(\frac{1}{1-1 / 4}ight) \ & =9 	imes 10^{9} 	imes 4 	imes 10^{-6} 	imes \frac{4}{3} \mathrm{NC} \end{aligned} $$ Note : $1+\frac{1}{4}+\frac{1}{16}+\ldots$ is a GP whose sum is given by $S=\frac{a}{1-r}$ where a is the first term (1 here) and $r$ is the common ratio ( $1 / 4$ here)\