Question

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.00mm away. What is the initial acceleration of the proton after it is released?

Answer 1

F = m.a
   = mass x acceleration

So,
a = F / m
Where,
F is coulomb repulsion.
k q q / r^2
with,
k = 9 x 10^9 in SI units (q in C and r in m)

Hence,
a = 2.40 × 10^11 m/s^2