A particle of mass 89 g and charge 20 µC is released from rest when it is 47 cm from a second particle of charge −28 µC. Determine the magnitude of the initial acceleration of the 89 g particle. Answer in units of m/s 2 .

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

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

i'm thinking more like energy

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