Question

Two charges $2\mu C$ and $1\mu C$ are placed at a distance of \[10cm\]. The distance of third charge from one of the charges so that it does not experience any force is:
\[\begin{align}
  & \text{A}\text{. 4}\text{.14cm from 2}\mu \text{C } \\
 & \text{B}\text{. 5}\text{.858cm from 2}\mu \text{C} \\
 & \text{C}\text{. 3}\text{.52cm from 2}\mu \text{C} \\
 & \text{A}\text{. 6}\text{.48cm from 2}\mu \text{C} \\
\end{align}\]

Answer 1

Hint: We know that the electric force experienced due to a pair charges is given by Coulomb's force. Clearly, the electric force depends on the charge $q$ and inversely proportional to the square of the distance between them. Since it is given that the distance of third charge from one of the charges so that it does not experience any force is, we can equate the force due to $q_{1}, q_{2}$ on $q_{3}$ and hence find the distance of $q_{3}$ with respect to $2\mu C$.

Formula used:
$E=\dfrac{q_{1}q_{2}}{4\pi\epsilon_{0}r^{2}}$

Complete answer:
Given, that the charges $q_{1}=2\mu C=2\times 10^{-6}C$ and $q_{2}=1\mu C=1\times 10^{-6}C$, also their distance between them is $r=10cm$. Let us assume that the third point is of charge $q_{3}$ and a distance of $r$ from charge$q_{1}$, then it is at a distance $10-r$ distance from $q_{2}$ since, $q_{1}$ and $q_{2}$ are \[10cm\] apart.
Let the force on $q_{3}$ due to $q_{1}$ is given as $E_{q_{1}q_{3}}=\dfrac{k\times q_{1}\times q_{3}}{r_{13}^{2}}$
Similarly, the force on $q_{3}$ due to $q_{2}$ is given as $E_{q_{2}q_{3}}=\dfrac{k\times q_{2}\times q_{3}}{r_{23}^{2}}$

Then, since it is given that, the electric field at $q_{3}$ is zero, then we can say that the force on $q_{3}$ due to $q_{1}$ is cancelled by the force on $q_{3}$ due to $q_{2}$ or they are equal in magnitude and opposite in direction, then we can say, $E_{q_{1}q_{3}}+E_{q_{2}q_{3}}=0$
Or, $E_{q_{1}q_{3}}= -E_{q_{2}q_{3}}$
Let us only consider the magnitude, then $E_{q_{1}q_{3}}=E_{q_{2}q_{3}}$
Or, $\dfrac{k\times q_{1}\times q_{3}}{r_{13}^{2}}=\dfrac{k\times q_{2}\times q_{3}}{r_{23}^{2}}$
Substituting the values we get, $\dfrac{2\times 10^{-6}}{r^{2}}=\dfrac{1\times 10^{-6}}{(10-r)^{2}}$
$\Rightarrow$ $r^{2}=(10-r)^{2}\times 2$
$\Rightarrow$ $r=(10-r)\times\sqrt 2$
$\Rightarrow$ $r(1+\sqrt 2)= 10\sqrt 2$
$\Rightarrow$ $r=\dfrac{10\sqrt 2}{1+\sqrt 2}=\dfrac{14.14}{2.414}=5.857cm$
Hence the charge is at a distance \[5.858cm\] from the $2\mu C$ charge

So, the correct answer is “Option B”.

Note:
Here, since the option distance of $q_{3}$ is given in terms of the $2\mu C$ charge. We are assuming the distance between $q_{3}$ and $q_{1}=2\mu C$ as $r$. Conversely, you can take the distance between $q_{3}$ and $q_{2}=1\mu C$ as $r$. Then, we get the distance between $q_{3}$ and $q_{1}=2\mu C$ as $10-r$.