Answer
$q=70.8\times 10^{-12}\ C = 70.8\ pC$
Work Step by Step
Given
acceleration of the first particle $a_1 = 7\ m/s^2 $
acceleration of the second particle $a_2 = 9\ m/s^2 $
mass of first particle $m_1=6.3\times 10^-7\ kg$
distance between the two charges d = $3.2\times 10^{-3}\ m$
Magnitude of force acting on each particle is
$F= m_1a_1 = m_2a_2$
$F = (6.3\times 10^{-7}kg)(7m/s^2)=44.1\times 10^{-7}\ N$
According to coulombs law, force of attraction between two charges is given by:
$F = \frac{kq_1q_2}{d^2}$
$44.1\times 10^{-7}=\frac{9\times 10^{9}(q)(q)}{(3.2\times 10^{-3})^2}$
$q^2 = \frac{(44.1\times 10^{-7})(3.2\times 10^{-3})^2}{9\times 10^{9}}$
$q =\sqrt\frac{(44.1\times 10^{-7})(3.2\times 10^{-3})^2}{9\times 10^{9}}$
$q=7.1\ \times 10^{-11} C = 7.1\ \times 10^{-11} C$
