# 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

**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?

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

Let the two charges be q_{1} and -q_{2}

in the first case,

F = -kq_{1}q_{2} / r^{2}

here,

r = 1m

so,

F = -kq_{1}q_{2 }/1^{2}

or

F = -kq_{1}q_{2} .................(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 = (q_{1} + q_{2})/2

thus, the coulombic force will be

F' = k((q_{1} + q_{2})/2).((q_{1} + q_{2})/2) / 1^{2}

or

F' = F/3 = k(q_{1} + q_{2})/2)^{2}

or

F/3 = k(q_{1} + q_{2})/2)^{2}

or

F = 3k((q_{1} + q_{2})/2)^{2}..............................(2)

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

-kq_{1}q_{2} = 3k((q_{1} + q_{2})/2)^{2}

or by solving further we get

3[(q_{1} + q_{2})/2]^{2} = - q_{1}q_{2}

or

3[q_{1} ^{2} + q_{2} ^{2} + 2^{ }q_{1}q_{2}] / 4 = -q_{1}q_{2}

or

3[q_{1} ^{2} + q_{2} ^{2} + 2^{ }q_{1}q_{2} ] = -4q_{1}q_{2}

or

3q_{1} ^{2} + 3q_{2} ^{2} + 10q_{1}q_{2} ] = 0

now, by diving both sides by q_{2} ^{2}, we get

3(q_{1}/q_{2})^{2} + 10(q_{1}/q_{2}) + 3 = 0

let q_{1}/q_{2} = x

so, we have

3x^{2} + 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

q_{1}/q_{2} = x = 0.33

or

q_{1}/q_{2} = x = 3

Therefore the ratio is 3:1

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

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