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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
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B
-2q
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C
q2
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D
4q
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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 correct

2102255_332365_ans_fe8586f53e10410d9ef904b8b08227dc.png

Physics

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