Two charges 4q and q are placed at a distance l apart. An another charged particle Q is placed in between them(at mid-point) if resultant force on q is zero then the value of Q is
a. q
b. -q
c. 2q
d. -2q
 

Dear Student,

Please find below the solution to the asked query:

The situation will be as shown in the below diagram.


So, here the force due to +4q on +q is,

$F_{q,4q}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(4q\right)\left(q\right)}{l^2}$
Force due to Q on q is,

$F_{q,Q}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(Q\right)\left(q\right)}{{\left({\displaystyle \raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2}$
If the resultant force is zero, then the force due to Q should be opposite to the force due to +4q​. Therefore, the charge Q should be negative.

Magnitudes of the two forces should be same. So,

$$\begin{gathered} F_{q,4q}=F_{q,Q}\Rightarrow \frac{1}{4\pi {\epsilon }_0}\frac{\left(4q\right)\left(q\right)}{l^2}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(Q\right)\left(q\right)}{{\left({\displaystyle \raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2} \\ \Rightarrow 4q^2=4Qq \Rightarrow Q=q \end{gathered}$$

Therefore,

$Q=-q$

So, option (b) is correct.
 

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards