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 0.321 kg are dropped from a height of 2.15 m , but one of the balls is positively charged with q1 = 360 μC , and the second is negatively charged with q2=-360 μC .
Use conservation of energy to determine the difference in the speeds of the two balls when they hit the ground. (Neglect air resistance.)
v+−v− =

Answer 1

The difference in the speeds of the two balls when they hit the ground is mathematically given as

V=0.10m/s

What is the difference in the speeds of the two balls when they hit the ground.?

Generally, the equation for Acceleration due to  electric field is mathematically given as

$A=g+\frac{q1E}{m}\\\\Therefore\\\\A=9.8*\frac{600*1000*150*}{0.520}\\\\A=9.97m/s2$

For Negative charge

$A=g+\frac{q1E}{m}\\\\A=9.8*\frac{-600*1000*150*}{0.520}\\\\A=9.62m/s2$

In conclusion, the speed of the negative ball

$v1=\sqrt{2ah}\\\\v1=\sqrt{2*9.67*2m}$

v1=6.21m/s

The  difference in v

V=v1-v2

V=6.31-6.21

V=0.10m/s

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