Question

can anyone help me in this question please A particle of mass 27 g and charge 32 µC is
released from rest when it is 79 cm from a
second particle of charge −22 µC.
Determine the magnitude of the initial acceleration
of the 27 g particle.
Answer in units of m/s
2
.

Answer 1

F = m a = k Q1 Q2/d^2
so
a = F/m

m = 27 g = .027 kg

Q1 = 32*10^-6
Q2 = -22*10^-6
so F is attractive between them

F=(1/.027)k (32)(22)(10^-12)/.79^2

Answer 2

Oh
k is about 9*10^9