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
[tex]F=\frac{Kq_{A}q_{B}}{d^{2}}[/tex]
By substituting the values, we get
[tex]5.4 \times 10^{-4}=\frac{9 \times 10^{9}q_{A}q_{B}}{0.05^{2}}[/tex]
qA x qB = 1.5 x 10^-16 C
So, [tex]q_{A}\left ( 25 \times 10^{-9}-q_{A} \right )=1.5 \times 10^{-16}[/tex]
[tex]q_{A}^{2}-25 \times 10^{-9}q_{A}+1.5 \times 10^{-16}=0[/tex]
[tex]q_{A}=\frac{25\times 10^{-9} \pm \sqrt{6.25\times 10^{-16}-6 \times10^{-16}}}{2}[/tex]
[tex]q_{A}=\frac {25\times 10^{-9} \pm 5 \times 10^{-9}}}{2}[/tex]
qA = 15nC or 10nC
So, qB = 10 nC or 15 nC
