Question

Two insulated identically sized charged copper spheres A and B have their centers separated by a distance of 50 cm. Charges on each sphere is
`q=6.5xx10^(-7)C`
. A third sphere of the same size but uncharged is brought in contact with the first, then in contact with the second and finally removed from both. What is the new force of repulsion between A and B?

Answer 1

Here, Charge on `A = 6.5xx10^(-7) C` , charge on `B = 6.5xx10^(-7) C`
Their sizes are equal . When third sphere C of same size is brought in contact with B carrying `6.5xx10^(-7)` coulomb charge. As their sizes are equal, therefore , charge on each of the spheres B and C becomes
`q_(2) = ((6.5+3.25) xx10^(-7))/(2) C = 4.875xx10^(-7) C`
As `F = (1)/(4pi in_(0)) (q_(1) q_(2))/(r^(2)) , r = 0.5m :. F = (9xx10^(9)xx3.25xx10^(-7)xx4.875xx10^(-7))/((0.5)^(2)) = 5.7xx10^(-3) N`