Question

Three charges $+4q,Q$ and $+q$ are placed in a straight line with $Q$ midway in between the other two charges. If net force on $+q$ is zero, the charge $Q$ must be:

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

Answer 1

The correct option is A $-q$

Force on q due to 4q is $F_1=k\frac{4q^2}{(2r{)}^2}=k\frac{q^2}{r^2}$

Force on q due to Q is $F_2=k\frac{Qq}{r^2}$

As net force on $q$ is zero, so using superposition principle, $F_1+F_2=0$

or,$k\frac{q^2}{r^2}+k\frac{Qq}{r^2}=0$

or,$Qq=-q^2\Rightarrow Q=-q$