Question

Three charges $+4q,Q$ and $q$ are placed in a straight line of length $l$ at points $0,\frac{l}{2}$ and $l$ distance way from one end respectively. What should be $Q$ 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 $-q$

Given three charges are placed along a straight line as shown in the figure below,


For the net force acting on $q$ to be zero, $${\overrightarrow{F}}_{net}=\overrightarrow{F_1}+\overrightarrow{F_2}=0$$ Where , $\overrightarrow{F_1}\to$ Force between charges $4q$ and $q$.

$\overrightarrow{F_2}\to$ Force between charges $Q$ and $q$.

Thus, ${\overrightarrow{F}}_1=-{\overrightarrow{F}}_2$

Forces acting on $q$ are equal in magnitude but opposite in direction.

$\Rightarrow \frac{4q^2}{4\pi {\epsilon }_0{l}^3}\times l\hat{i}=-\frac{Qq}{4\pi {\epsilon }_0{\left(\frac{l}{2}\right)}^3}\times \left(\frac{l}{2}\right)\hat{i}$

$\therefore Q=-q$

Hence, option (a) is the correct answer.

Why this question? This questions provides a clear understanding to a student , regarding the application of concept of equilibrium. Recapitulate: According to coulomb's law of electrostatics, Force acting between any two charges $$F\propto q_1{q}_2$$ and if $r$ is the seperation distance between the charges then, $$F\propto \frac{\overrightarrow{r}}{r^3}$$