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

what do you think john? do you have any idea about it?

Answer 2

To be honest, I got sick and wasn't able to join the discussion:( That's why I need help badly

Answer 3

but u must have SOME lead .. or some idea.. something atleast right? :P

Answer 4

F = m a = mass x acceleration so a = F / m

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/)

Answer 5

and here i was trying to teach someone :P

Answer 6

I got the first and second part already but how to get Calculate the proton's speed after 1.50μs in the field, assuming it starts from rest.. I know for sure v=at but for some reason the answer is wrong when I used the answer in a

Answer 7

U can't use that.. cause the acceleration is not a constant.. ! u need to either

a) do an integral ..

or b) use energy conservation .. see how much potential energy it had, and how much of it gets converted to kinetic energy!

Answer 8

a = 2.40×1011 m/s2
that's my answer

Answer 9

Sorry, wrong advice. F= m a, but a is not constant here. Should have used conservation of energy and change in potential energy and probably have let the student work out more of it himself. Hard to know where to teach and where to answer. Giving wrong advice is least valuable!