Question

Three charges + 4q, Q and q are placed on a straight line of length l at points distance 0, $\left(\frac{1}{2}\right)$ and l respectively. What should be Q in order to make the net force on q zero?

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

Answer 1

The correct option is A - q

Force on q due to charge +4q which is at a distance l
$F_1=\frac{1}{4\pi {ϵ}_0}\times \frac{qQ}{{\left(\frac{1}{2}\right)}^2}$
Hence $F_1+F_2=0$
$\frac{1}{4\pi {ϵ}_0}\times \frac{4q\times q}{l^2}+\frac{1}{4\pi {ϵ}_0}\frac{qQ}{{\left(\frac{1}{2}\right)}^2}=0$
$\therefore Q=-q$