Asmall metal bead, labeled a, has a charge of 28 nc .it is touched to metal bead b, initially neutral, so that the two beads share the 28 nc charge, but not necessarily equally. when the two beads are then placed 5.0 cmapart, the force between them is 4.8×10^−4 n . assume that a has a greater charge. what is the charge qa and qb on the beads?

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} =

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

are there options? cause you ain’t. real specific

i think if it were me i whould divide 300 dide 15

step-by-step explanation: