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})-
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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

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