gneill
Mentor
How do I find the acceleration then, I'm missing the force
You don't need to find a particular value for the force. You just need to find an expression for it, and note what values it depends on, and whether or not any of them differ from case to another in your problem.

You don't need to find a particular value for the force. You just need to find an expression for it, and note what values it depends on, and whether or not any of them differ from case to another in your problem.
What differs in the problem is their distance maybe, their velocities and their direction. Other than that I assume their acceleration since thats what we are looking for.

gneill
Mentor
What differs in the problem is their distance maybe, their velocities and their direction. Other than that I assume their acceleration since thats what we are looking for.
That's why you need to write an expression for the acceleration, so you can test your assumptions.

That's why you need to write an expression for the acceleration, so you can test your assumptions.
So like Acceleration,=??? meters/s^2 Q=1.6x10^-19 C, m=1.67x10^-27kg, E=?????? N/C, F=???? Newtons,
a=F/m
F= QxE
a=(QxE)/m
Only problem is there is nothing about direction or velocity

gneill
Mentor
So like Acceleration,=??? meters/s^2 Q=1.6x10^-19 C, m=1.67x10^-27kg, E=?????? N/C, F=???? Newtons,
a=F/m
F= QxE
a=(QxE)/m
Only problem is there is nothing about direction or velocity
That's not a problem, it's a bonus and a clear result! You now now that the acceleration does not depend upon direction or speed or distance; It only depends on the charge, the mass, and the electric field. So what can you conclude for the different cases presented?

That's not a problem, it's a bonus and a clear result! You now now that the acceleration does not depend upon direction or speed or distance; It only depends on the charge, the mass, and the electric field. So what can you conclude for the different cases presented?
Their accelerations are the same?? That just doesn't seem like him. (professor)

gneill
Mentor
Their accelerations are the same?? That just doesn't seem like him. (professor)
Yes, the accelerations are all identical. It is the correct result for the question in the form it was given.

The question may seem like a trick, or perhaps it is incorrect due to a typo or otherwise, yet even so it got you to think about the physics of the situation.

Ok thanks gneill and berkeman. I guess you guys had said their accelerations were same in the beginning but it definitely helped figuring out why. I''ll talk to my professor tomorrow and I think it's actually due Tuesday so if theres something I'm missing I'll bring it up tomorrow?

Thanks!!

gneill
Mentor
Ok thanks gneill and berkeman. I guess you guys had said their accelerations were same in the beginning but it definitely helped figuring out why. I''ll talk to my professor tomorrow and I think it's actually due Tuesday so if theres something I'm missing I'll bring it up tomorrow?
That would be fine!
Thanks!!
You're very welcome.