Ask a question
0 0

Physics speed

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 20Ca40)
Tutors, please sign in to answer this question.

2 Answers

Assuming initial gvelocity of Ca ++ is zero
Final velocity = initial velocity  +  (acceleration x time)
acceleration = Forec/mass
mass of Ca is 40 amu = 40 x 1.66 x 10-27 Kg
Force acting on Ca++ = 2 x 1.6 x 10-19 x 200 = 640 x 10-19 N
acceleration of Ca++ = (640 x 10-19)/ (40 x 1.66 x 10-27) = 9.6 x 108 m/s2
Final velocity = 0 + (9.6 x 108 x 50 x 10-3 m/s) = 480 x 105 m/s = 48,000 Km/s
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