Question

Two point charges $4\mu C$ and $1\mu C$ are placed $9m$ apart. A third point charge $q$ is placed on line joining $4\mu C$ and $1\mu C$. Value of $q$, so that all three charges can remain in equilibrium is

• A. $-2\mu C$
• B. $-\frac{4}{9}\mu C$
• C. $-\frac{3}{8}\mu C$
• D. $-\frac{5}{9}\mu C$

Answer 1

The correct option is B $-\frac{4}{9}\mu C$

The charge $q$ should be negative in nature for keeping the three charges in equilibrium. Also the charge $q$ should be placed between the given charges.
So, only magnitude of charge $q$ is taken into consideration for calculations below:

Here, $x$ is the distance of $q$ from $1\mu C$ charge in meters.

$\Rightarrow \frac{K\times 4\times q}{(9-x{)}^2}=\frac{K\times 1\times q}{x^2}$
$4x^2=(9-x{)}^2\Rightarrow x=+3or-9$

For, $x=-9$ the equilibrium cannot be maintained.

For, $x=+3$
$\frac{K\times 4\times 1}{9^2}=\frac{K\times q\times 1}{x^2}\Rightarrow q=4(\frac{x}{9}{)}^{2}q=\frac{4}{9}\mu C$

$\Rightarrow q=-\frac{4}{9}\mu C$