A calcium ion Ca++ enters with zero velocity a region of uniform electric field of 200N/C pointing east. What is its speed after 50ms? (Symbol of calcium atom is
_{20}Ca^{40})
Hi David
The force on the ion is q*E where q is the charge and E is the magnitude of the electric field. I this case it is a constant force so the acceleration is constant. Recalling the equation for velocity of a particle undergoing a constant acceleration
and starting from rest i.e. v=a*t now a=qE/m and t=50msec m= the mass of the ion.
So v=t*q*E/m you will have to do the arithmetic
Regards
Jim