THE CENTRES OF 2 IDENTICAL SMALL CONDUCTING SPHERES ARE 1M APART.THEY CARRY CHARGES OF OPPOSITE KIND AND ATTRACT EACH OTHER WITH A FORCE F.WHEN THEY ARE CONNECTED BY A CONDUCTING WIRE THEY REPEL EACH OTHER WITH A FORCE F/3.WHAT IS THE RATIO OF MAGNITUDE OF CHARGES CARRIED BY THECHARGES INITIALLY?

 

(A)1:1    (B)2:1   (C)3:1   (D)4:1

Let the two charges be q₁ and -q₂

in the first case,

F = -kq₁q₂ / r²

here,

r = 1m

so,

F = -kq₁q₂ /1²

or

F = -kq₁q₂ .................(1)

and

when the spheres are connected by wire the charge transfer takes place. It is such that the charge now is equally distributed over the two spheres

so,

charge on each sphere will be = (q₁ + q₂)/2

thus, the coulombic force will be

F' = k((q₁ + q₂)/2).((q₁ + q₂)/2) / 1²

or

F' = F/3 = k(q₁ + q₂)/2)²

or

F/3 = k(q₁ + q₂)/2)²

or

F = 3k((q₁ + q₂)/2)²..............................(2)

now, by equation (1) and (2), we get

-kq₁q₂ = 3k((q₁ + q₂)/2)²

or by solving further we get

3[(q₁ + q₂)/2]² = - q₁q₂

or

3[q₁ ² + q₂ ² + 2 q₁q₂] / 4 = -q₁q₂

or

3[q₁ ² + q₂ ² + 2 q₁q₂ ] = -4q₁q₂

or

3q₁ ² + 3q₂ ² + 10q₁q₂ ] = 0

now, by diving both sides by q₂ ², we get

3(q₁/q₂)² + 10(q₁/q₂) + 3 = 0

let q₁/q₂ = x

so, we have

3x² + 10x + 3 = 0

which is a form of a quadratic equation, the roots of which are

x = -0.33 and -3

in magnitude

|x| = 0.33 or 3

thus, the ratio intial of charges will be

q₁/q₂ = x = 0.33

or

q₁/q₂ = x = 3

Therefore the ratio is 3:1