Correct Answer - `+-5.6xx10^(-7)C`
In Fig, A and B are two balloons attached to weight W by strings `OA = OB = 1m`.
`AB = 0.6m`, C is centre of AB.
`:. AC = BC = 0.3m`.
`OC = sqrt(OA^(2) - AC^(2)) = sqrt(1^(2) - 0.3^(2)) = 0.954`
Force of repulsion between flost in equilibrium,
`:.` Using triangle law of vactors, we get
`(F)/(AC) = (w)/(OC)`,
where w = upthrust on each balloon.
As w + w = W,
`:. w = W//2`
`:. F = (AC)/(OC) xx (W)/(2)`
`(q xx q)/(4pi in_(0) (AB)^(2)) = (AC)/(OC) xx (mg)/(2)`
`q^(2) = 4pi in_(0) (AB)^(2) xx (AC)/(OC) xx (mg)/(2)`
`= ((0.6)^(2)xx0.3xx(5xx10^(-3)xx9.8))/(9xx10^(9)xx0.954xx2)`
`q^(2) = (0.36xx0.3xx49)/(9xx0.954xx2) xx 10^(-12)`
`q = +- 0.56xx10^(-6)C = +-5.6xx10^(-7)C`