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
q2
D
4q

Solution

## The correct option is A -q$$\textbf{Step 1: Force on q due to all other charges [Ref Fig 1]}$$ 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 :$$F_1=K\dfrac { qQ }{ (l/2)^{ 2 } } =\dfrac { 4KqQ }{ l^{ 2 } }$$                                                   $$....(1)$$.The force exerted on the charge q due to 4q :$$F_2=K\dfrac { q\ 4q} { l^{ 2 } } =\dfrac { 4Kq^{ 2 } }{ l^{ 2 } }$$                                                       $$....(2)$$.$$\textbf{Step 2 : Net force on q }$$Net force $$= F_1+ F_2 = 0$$  Using  equations (1) and (2)$$\dfrac { 4KqQ }{ l^{ 2 } } +\dfrac { 4Kq^{ 2 } }{ l^{ 2 } } = 0$$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 correctPhysics

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