Question

A small metal bead, labeled A has a charge of 25 nC. It is touched to metal bead B, initially neutral, so that the two beads share the 25 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cm apart, the force between them is 5.4 x 10e-4 N. What are the charges qa and qb on the beads?

Answer 1

Answer:

15 nC and 10 nC

Explanation:

qA + qB = 25 nC = 25 x 10^-9 C

F = 5.4 x 10^-4 N

d = 5 cm = 0.05 m

Use the Coulomb's law

$F=\frac{Kq_{A}q_{B}}{d^{2}}$

By substituting the values, we get

$5.4 \times 10^{-4}=\frac{9 \times 10^{9}q_{A}q_{B}}{0.05^{2}}$

qA x qB = 1.5 x 10^-16 C

So, $q_{A}\left ( 25 \times 10^{-9}-q_{A} \right )=1.5 \times 10^{-16}$

$q_{A}^{2}-25 \times 10^{-9}q_{A}+1.5 \times 10^{-16}=0$

$q_{A}=\frac{25\times 10^{-9} \pm \sqrt{6.25\times 10^{-16}-6 \times10^{-16}}}{2}$

$q_{A}=\frac {25\times 10^{-9} \pm 5 \times 10^{-9}}}{2}$

qA = 15nC or 10nC

So, qB = 10 nC or 15 nC