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Posted by on Sunday, February 7, 2010 at 6:56pm.

Near the surface of the Earth there is an electric field of about V=150 V/m which points downward. Two identical balls with mass m= 0.550 kg are dropped from a height of 2.30 m, but one of the balls is positively charged with q1= 800 microCoulombs , and the second is negatively charged with q2= -800 microCoulombs.
Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)

  • Physics - , Sunday, February 7, 2010 at 7:12pm

    Neglecting air resisistance, of course, which would be a very significant factor as compared to E.

    Energy given up: mgh+Eqh


    1/2 mv^2=mgh+Eqh
    Now pay attention to charge, the + charge will go faster, the negative charge will slow.

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