Answer:
a = 256.36 m/s²
Explanation:
given,
Charge of the 89 g particle = 20µC
distance from the second particle = 47 cm = 0.47 m
charge of the second particle = -28 µC
acceleration of 89 g particle = ?
Using Coulomb force formula
[tex]F = \dfrac{kQ_1Q_2}{r^2}[/tex]
[tex]F = \dfrac{9\times 10^9\times 20\times 10^{-6}\times -28\times 10^{-6}}{0.47^2}[/tex]
F = 22.82 N
For acceleration calculation
Using 2nd law of force
F = m a
22.82 = 0.089 a
a = 256.36 m/s²
Hence, the acceleration of the particle is equal to a = 256.36 m/s²
