Question

Please tell the answer​

Answer 1

Answer:

The answer is option (B) - Q/4 (1+2√2)

Explanation:

Here given that the system is in equilibrium,hence the force exerted by each of these particles are cancelled out.

Now consider the first point Q, and equate the opposite forces acting,since they are equal and cancels out.

Look at the attached image file and there are 3 forces acting on the charge Q, i.e F1 ,F2,F3

F1 and F2 are acting perpendicular to each other hence the net charge is in the direction of F3, and since the system is in equilibrium

Net resultant force due to F1 and F2= -F3 (Not F3, Its  -F3 coz the force should cancel out the resultant charge)

k constant is on both sides hence can be cancelled out, now remaining equation

F1 + F2= -(F3)

F1 = F2 =$\frac{Q^{2} }{r^{2} }$

Now force of F3,

Here on charge Q there are two force acting on it,

One:Due to the charge q in the centre.(Here distance b/w charge is$\frac{\sqrt{2} r}{2}$ )

Two: Another charge Q on the diagonal side.(Here the distance is $\sqrt{2} r$)

F3= $\frac{Qq}{\frac{( \sqrt{2}r}{(2})^{2} } + \frac{Q^{2} }{(\sqrt{2}r)^{2} }$   (value shown in the image attached)

Equating

F1+F2= -(F3)  

resultant of F1 and F2 since perpendicular to each other is

resultant F1+F2= $\sqrt{F1^{2} + F2^{2} } }$

which after simplification will be equal to

=$\frac{\sqrt{2}Q^{2} }{r^{2} }$

Now equating resultant of F1+F2 and F3 gives

$\frac{\sqrt{2}Q^{2}  }{r^{2} }$ = $\frac{Qq}{\frac{( \sqrt{2}r}{(2})^{2}   } + \frac{Q^{2} }{(\sqrt{2}r)^{2}  }$

Further canceling and keeping q in LHS and all other variables in RHS , we get

q=Q/4 (1+2√2)

But since q should be of opposite charge so that the resultant force is nullified,

Hence q= - Q/4 (1+2√2) Option (B)