Question # 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

# 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 )

Solution

( 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.