Two equally charged particles, held 3.2 x 10^–3 m apart, are released from rest. The initial acceleration of the first particle is observed to be 7.0 m/s² and that of the second to be 9.0 m/s². If the mass of the first particle is 6.3 x 10^–7 kg, what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?

Solution

Given,

Distance between the charges, r = $3.2 \times 10^{- 3} m$
$Iinitial acceleration of first particle, a_{1} = 7.0 m / s^{2} Initial acceleration of second particle, a_{2} = 9.0 m / s^{2} Mass of first particle, m_{1} = 6.3 \times 10^{- 7} kg, Mas of second particle, m_{2} = ?$

a) Since, $F_{1} = F_{2}$

$∴$    $m_{1} a_{1} = m_{2} a_{2}$

$mass of second particle - m_{2} = \frac{m_{1} a_{1}}{a_{2}} = \frac{6.3 \times 10^{- 7} \times 7.0}{9.0} = 4.9 \times 10^{- 7} kg$

b)As,
   $F_{1} = F_{2} = \frac{q_{1} q_{2}}{4 {πε}_{0} r^{2}} = m_{1} a_{1} = 6.3 \times 10^{- 7} \times 7.0 = 44.1 \times 10^{- 7}$

$∴$ $\frac{9 \times 10^{9} q^{2}}{{(3.2 \times 10^{- 3})}^{2}} = 44.1 \times 10^{- 7}$

$∴$    $q = 7.1 \times 10^{- 11} C .$

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