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Near the surface of the Earth there is an electric field of about 151 V/m which points downward. Two identical balls with mass m = 0.568 kg are dropped from a height of 1.70 m, but one of the balls is positively charged with q1 = 568 µC, and the second is negatively charged with q2 = -568 µC. 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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Final answer:

The difference in speed when the two balls hit the ground can be found using conservation of energy, accounting for the electric force exerted on the charged balls. It is determined by the work done by gravity and the respective electric forces on both the positively and negatively charged balls.

Step-by-step explanation:

To determine the difference in speed when the two balls hit the ground, we consider the effects of both gravity and the electric field. Initially, both balls have the same potential energy due to their height. As they fall, this potential energy is converted into kinetic energy. The positively charged ball experiences an upward electric force, while the negatively charged one experiences a downward force. This force does work on the balls, affecting their kinetic energy when they reach the ground.

The work done by gravity is Wg = mgh, and by the electric field is We = qEh for each ball. The total change in kinetic energy is equal to the work done by both forces. For the positively charged ball, the net work done is W_net_pos = mgh - qEh, since the electric force opposes gravity. For the negatively charged ball, it is W_net_neg = mgh + qEh.

Using the conservation of energy and setting the initial kinetic energy to zero (since they are dropped), we can equate the net work to the final kinetic energy:

  • For the positively charged ball: ½mv_pos^2 = mgh - qEh
  • For the negatively charged ball: ½mv_neg^2 = mgh + qEh

Subtracting these equations yields the difference in kinetic energy, and taking the square root gives the difference in speed:

Δv = √(2qEh/m)

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