Answer
$a_A=45\frac{m}{s^2}$
$a_B=22.5\frac{m}{s^2}$
Work Step by Step
We know that
$F=K\frac{q^2}{r^2}$
We plug in the known values to obtain:
$F=9\times 10^9\frac{(5\times 10^{-6})^2}{(1.00)^2}$
$F=0.225N$
Now $a_A=\frac{F}{m_A}$
We plug in the known values to obtain:
$a_A=\frac{0.225}{5\times 10^{-3}}$
$a_A=45\frac{m}{s^2}$
Also, $a_B=\frac{F}{m_B}$
We plug in the known values to obtain:
$a_B=\frac{0.225}{10\times 10^{-3}}$
$a_B=22.5\frac{m}{s^2}$
