Question

Three charges $+4q,Q$ and $+q$ are placed in a straight line such that the charge $Q$ is equidistant from both the other charges. What should be the charge $Q$ in order to make the net force on $+q$ to be zero ?

• A. $-q$
• B. $4q$
• C. $-\frac{q}{2}$
• D. $-2q$

Answer 1

The correct option is A $-q$

Let the distance between the charges $+4q$ and $+q$ be $l$.


For $q$ to be in equilibrium $(F_{\text{net}}=0),Q$ and $q$ must be oppositely charged.
Let $F_1$ and $F_2$ is the force experienced by $q$ due to charges $Q$ and $+4q$ respectively.
Thus, $|F_1|=|F_2|$
$\Rightarrow \frac{kQq}{{\left(\frac{l}{2}\right)}^2}=\frac{k\left(4q\right)\left(q\right)}{l^2}$
($F_1$ is the attractive force by $Q$ and $F_2$ is repulsive force by $+4q$)
$\Rightarrow \frac{4kQq}{l^2}=\frac{4k{q}^2}{l^2}$
$\Rightarrow Q=q$
Since $Q$ and $q$ are of opposite nature.
$\therefore Q=-q$
Hence, option (a) is correct.

$\begin{array}{c}\text{It intends to test application of Coulomb's}\\ \text{law to achieve equilibrium of a charged body.}\end{array}$