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Near the surface of the Earth there is an electric field of about V=150 V/m which points downward. Two identical balls with mas...Asked by Ambrie
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.)
Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)
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Answered by
bobpursley
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.
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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