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 = 0.680 kg are dropped from a height of 1.80 m , but one of the balls is positively charged with q1 = 850 μC , and the second is negatively charged withq2 = -850 μC

Part A

Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)

Answer 1

If there were no electric field, two equally-massive balls dropped from equal heights would acquire the same amount of kinetic energy (KE) and hit the ground at the same speed.

An electric field will add or subtract energy from each ball, depending on the charge of the ball, the magnitude and orientation of the field, and the distance the ball travels through the field. This field has a magnitude of 150 V/m, and since each ball is falling 1.80 meters through the field, that's a potential difference of 270 volts between the top of the ball's fall and the bottom. The field is oriented downward, and the direction of a field is always the direction of force on a positive charge. Therefore, the positively charged ball will be pushed downward and have energy ADDED to its falling KE. The negatively charged ball will be pushed upward and have energy SUBTRACTED from its falling KE.

The energy gained or lost by a charge is the product of the voltage difference and the magnitude of the charge. Both have the same magnitude of charge = 850 uC or 8.5*10^(-4) C.

delta E = V*q = (270V)(8.5*10^(-4) C) = 0.2295 J

The positively charged ball will gain 0.2295 J of KE, while the negative ball will lose 0.2295 J of KE. They will hit the

ground with a difference of 0.459 J. Since they have the same mass, we can determine the difference in speed

delta K = 1/2 m(delta v)^2

0.459 J = 1/2*(0.680 kg)(delta v)^2

0.459 J = (0.340 kg)(delta v)^2

(delta v)^2 = 1.35 m^2/s^2

delta v = 1.16 m/s