Question

Two charges $4q$ and $q$ are placed at a distance $\ell$ 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. $-2q$

Answer 1

The correct option is B $-q$

The force due to $+4q$ on $+q$ is given as,

$F_1=\frac{1}{4\pi {\epsilon }_0}\times \frac{4q\times q}{l^2}$

The force due to $Q$ on $+q$ is given as,

$F_2=\frac{1}{4\pi {\epsilon }_0}\times \frac{Q\times q}{{\left(\frac{l}{2}\right)}^2}$

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.

The magnitudes of the two forces should be the same and it can be written as,

$\frac{1}{4\pi {\epsilon }_0}\times \frac{4q\times q}{l^2}=\frac{1}{4\pi {\epsilon }_0}\times \frac{Q\times q}{{\left(\frac{l}{2}\right)}^2}$

$Q=q$

Thus, the value of $Q$ is $-q$.