Question

Two point charges $+9e$ and $+e$ are kept $16cm$ apart to each other. Where a third charge $q$ is placed between them so that the system is in equilibrium state:
A) $24cm$ from $+9e$
B) $12cm$ from $+9e$
C) $24cm$ from $+e$
D) $12cm$ from $+e$

Answer 1

Hint: suppose the charge $q$ is placed at a distance $r$ from the $ + 9e$ charge. For the system to be in equilibrium, the force between $ + 9e$ charge and $q$ charge must be equal to the force between $ + e$ charge and $q$ charge. Equate these two forces and solve for $r$ to get the answer.

Formula Used:
Force between two point charges, $F = \dfrac{{KQq}}{{{r^2}}}$ where, $K$ is the proportionality constant, $Q$ is magnitude of charge 1, $q$ is the magnitude of charge 2, and $r$ is the distance between the charges.

Complete Step by Step Solution:
Suppose the charge $q$ is placed between the two given charges at a distance of $r$ from the charge $ + 9e$ . This means that the distance of charge $q$ from charge $ + e$ will be $(16 - r)$ (given charges $ + 9e$ and $ + e$ are kept $16cm$ apart to each other)
Now let the force between the charges $ + 9e$ and $q$ be ${F_1}$
We get, ${F_1} = \dfrac{{K \times 9e \times q}}{{{r^2}}}$
Similarly, let the force between the charges $ + e$ and $q$ be ${F_2}$
We get, ${F_2} = \dfrac{{K \times e \times q}}{{{{(16 - r)}^2}}}$
Since these two forces must be equal for the system to be in an equilibrium state, we can write
$\Rightarrow$ $\dfrac{{K \times 9e \times q}}{{{r^2}}} = \dfrac{{K \times e \times q}}{{{{(16 - r)}^2}}}$
We get $\dfrac{9}{{{r^2}}} = \dfrac{1}{{{{(16 - r)}^2}}}$ (cancelling like terms)
Now, taking square root on both the sides, we get $\dfrac{3}{r} = \dfrac{1}{{16 - r}}$
Which gives $48 - 3r = r$ or, $48 = 4r \Rightarrow r = 12cm$
Therefore, the charge $q$ is at a distance of $12cm$ from charge $ + 9e$

Hence, Option (B) is the correct answer.

Note: In the formula for force between two charges, $K$ is the proportionality constant, often called the coulomb’s constant. Its value is $9 \times {10^9}N{m^2}{C^{ - 1}}$ . We did not put its value in this question as it was getting cancelled later on. Putting its value would have complicated the calculation. But this will not be the case every time. So, it is better to memorize its value for future use.