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

Asked by HARSHALI DOSHI | 16th Jul, 2014, 08:26: AM

Expert Answer:

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

Answered by Sheetal Jalan | 18th Jul, 2014, 01:20: PM