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 download. Two identical balls with mass m = 0.580 kg are dropped from a height of 2.10 m, but one of the balls is positively charged with q_1 = 950 mu C, and the second is negatively charged with q_2 = -950 mu C. Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.) Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Electric force on the 1st ball, Fe = +qE downward.

Total force downward, F1 = mg + qE

Acceleration of the ball downward,

a1 = F/m = g + qE/m

Simuilarly, acceleration of the 2nd ball,

a2 = g - qE/m

Velocity on hitting the ground,(initial velocity u = 0)

v12 = 2.a1.h

Similarly, v22 =2.a2.h

Substituting the given values,

qE/m = 950*10-6 *150/0.58

= 0.2457 m/s2

Hence, a1 = g + qE/m = 9.8 + 0.2457 = 10.0457 m/s2

And, a2 = g - qE/m = 9.8 - 0.2457 = 9.5543 m/s2

hence, v12 =2*a1*h = 4*a1 = 4*10.0457 = 40.1828

v1 = sqrt(40.1828) = 6.339 m/s

Similarly, v22 = 2*a2*h =4*a2 = 4*9.5543 = 38.217

v2 = sqrt(38.217) = 6.182 m/s

Therefore difference,

v1 - v2 = 6.339- 6.182 = 0.157 m/s

NOTE: The qn says use conservation ofenrgy.

At the top, Ug + Ue +K = mgh + qV(top)+0

At the bottom, Ug' + Ue' +K' = 0 + qv(ground)+0.5m.v2

Equating(from consrvation of energy,

0.5mv2 = mgh +q(V - V')

= mgh + q?V

= mgh + q.E.h

Therefore, v12 = 2.(g+qE/m)h = 2a1.h as above.

For q -ve, v22 = 2.(g- qE/m)h = 2a2.h

The rest of the solution is the same as shown above.

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