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})-
CREATE FREE ACCOUNT
- Access thousands of free resources from expert tutors
- Comment on posts and interact with the authors
- Ask questions and get free answers from tutors
- View videos, take interactive quizzes, and more!

- Become a Student
- Become a Student
- Sign In

Tutors, please sign in to answer this question.

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

Regards

Jim

Timothy R.

Mathematics and Physics Tutor

Edison, NJ

4.8
(9 ratings)

David M.

Experienced Teacher and Tutor for Math, Science and beyond!

New York, NY

5.0
(593 ratings)

Matt B.

Friendly, patient, and expert tutor

New York, NY

4.9
(266 ratings)