Near the surface of the Earth there is an electric field of about 150 V/m which

Question

Near the surface of the Earth there is an electric field of about 150 V/m which points downward. Two identical balls with mass m = 0.520 kg are dropped from a height of 2.00 m, but one of the balls is positively charged with q1 = 700 µC, and the second is negatively charged with q2 = -700 µC. Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)

Explanation / Answer

as direction of E is downward. hence we can treat the electric potential =0 at the ground. now initial total pot. energy of the 1 particle. = mgh + q1Eh = 10.4 + 0.21 final total pot. energy of the 1 particle = 0 + 0 + kinetic energy = .5 m v^2 by conservation of energy .5 m v^2 = 10.61 => v= 6.39 m/s initial total pot. energy of the 2 particle. = mgh + q2Eh = 10.4 - 0.21 final total pot. energy of the 2 particle = K.E. = 0.5mv^2 by conservation of energy .5 m v^2 = 10.19 => v=6.26 m/s hence difference between the two velocity is 0.13 m/s answer

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