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