Question

Three charges +4q, Q and q are placed in a straight line of length $l$ at points at distances 0, $l/2$, and $l$ respectively. What should be Q in order to make the net force on q to be zero?

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

Answer 1

The correct option is A -q

$\text{Step 1: Force on q due to all other charges [Ref Fig 1]}$

Considering the charge Q to be positive in the beginning, mention the direction of forces in the figure.

Applying Coulomb's law, the force exerted on the charge q due to Q :

$F_1=K\frac{qQ}{(l/2{)}^2}=\frac{4KqQ}{l^2}$ $....\left(1\right)$.

The force exerted on the charge q due to 4q :

$F_2=K\frac{q4q}{l^2}=\frac{4K{q}^2}{l^2}$ $....\left(2\right)$.

$\text{Step 2 : Net force on q}$

Net force $=F_1+F_2=0$

Using equations (1) and (2)

$\frac{4KqQ}{l^2}+\frac{4K{q}^2}{l^2}=0$

So, $qQ+q^2=0$

$\Rightarrow q(Q+q)=0$

Since $q$ cannot be 0

$\therefore Q+q=0$

$\Rightarrow Q=-q$
Hence, Q should be -q in order to make the net force on q to be zero.

Note -Now we know the nature of $Q$ we can show the actual direction of the forces and see that forces are in opposite directions making the net force on $q$ zero.[Ref Fig 2]

Hence, Option A is correct