Question

An infinite number of point charges, each carrying $1\mu C$ charge, are placed along the y-axis at $y=$$1m,2m,4m,8m........$ The total force on a $1C$ point charge, placed at the origin, is $X\times {10}^{3}N$. The value of $X$, to the nearest integer, is ……… [Take $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4\pi {\epsilon }_0=9\times {10}^9{\displaystyle \raisebox{1ex}{$N{m}^2$}\!\left/ \!\raisebox{-1ex}{$C^2$}\right.}$}\right.$]

Answer 1

Step 1: Given that.

Charge of $1\mu C$ are placed point on the $y-axis$ at $y=$$1m,2m,4m,8m........$respectively from the total force on a $1C$ point charge, placed at $X\times {10}^3$.As shown in the figure below.$\left[\frac{1}{4\pi {\epsilon }_0}=k=9\times {10}^9\raisebox{1ex}{$N{m}^2$}\!\left/ \!\raisebox{-1ex}{$C^2$}\right.\right]$

Step 2: Formula used.

Coulomb's Law:

$F=\frac{k{q}_1{q}_1}{r^2}$

Where $F$ is force, $q_1=q_2$is charge, and $r$ is the distance between charge.

Now,

The net force on $1C$ due to all each point charges of $1\mu C$.

$F_{net}=\frac{k{q}_1{q}_2}{r_1^2}+\frac{k{q}_1{q}_2}{r_2^2}+\frac{k{q}_1{q}_2}{r_3^2}+......$

$\Rightarrow$$F_{net}=k{q}_1{q}_2\left[\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+......\right]$

$\Rightarrow$$F_{net}=9\times {10}^9\times 1\times 1\times {10}^{-6}\times \left[1+\frac{1}{2}+{\left(\frac{1}{2^2}\right)}^2+......\right]$

$\Rightarrow$$F_{net}=9\times {10}^9\times {10}^{-6}\times \left[\frac{1}{1-{\displaystyle \frac{1}{4}}}\right]$

$\Rightarrow$$F_{net}=12\times {10}^3\hspace{0.17em}N$

Hence, the value of integer $X\times {10}^{3}N$ is $12$ as compared to $F_{net}$.