A 2.0 g ball with a +0.05 C charge is moved from the negative plate to the positive plate, and then is released. The potential difference between the plates is 12.0V. When the ball strikes the (–) plate its velocity is most nearly equal to

24.5 m/s

Explanation:

Since the work done by the electric field on the charge equals the kinetic energy of the ball, then

qV = 1/2mv²

and v = √(2qV/m)

where q = charge = + 0.05 C, V = potential difference = 12.0 V, m = mass of ball = 2.0 g = 0.002 kg

Substituting the values into v, we have

v = √(2 × + 0.05 C × 12.0 V/0.002 kg)

= √(1.2 CV/.002 kg)

= √(600 CV/kg)

= 24.5 m/s

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