Two charges 4q and q are placed at ...
Question
Two charges 4q and q are placed at a distance l apart. A third charged particle Q is placed at the middle of them. If resultant force on q is zero then the value of
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Two charges ( 4 mathrm{q} ) and ( mathrm{q} ) are placed at a distance ( ell ) apart. A third charged particle ( Q ) is placed at the middle of them. If resultant force on q is zero then
the value of ( Q ) is :
(1) ( mathrm{q} )
( (2)-q )
(3) ( 2 q )
(4) ( -2 q )

NEET/Medical Exams
Physics
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Two charges 4q and q are placed at a distance l apart. A third charged particle Q is placed at the middle of them. If resultant force on q is zero then the value of Q is
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( F_{q, 4 q}=frac{1}{4 pi varepsilon_{0}} frac{(4 q)(q)}{l^{2}} )
Force due to Q on ( q ) is,
( F_{q, Q}=frac{1}{4 pi varepsilon_{0}} frac{(Q)(q)}{(l / 2)^{2}} )
If the resultant force is zero, then the force due to ( Q ) should be opposite to the force due to ( +4 q ) Therefore, the charge ( Q ) should be negative.

Magnitudes of the two forces should be same. So
( F_{q, 4 q}=F_{q, Q} Rightarrow frac{1}{4 pi varepsilon_{0}} frac{(4 q)(q)}{l^{2}}=frac{1}{4 pi varepsilon_{0}} frac{(Q)(q)}{(l / 2)^{2}} )
( Rightarrow 4 q^{2}=4 Q q Rightarrow Q=q )
Therefore,
( Q=-q )
So, option (b) is correct.

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