In: Physics
Near the surface of the Earth there is an electric field of about 150 V/m which points downward. Two identical balls with mass 0.407kg are dropped from a height of 2.29m , but one of the balls is positively charged with q1 = 338?C , and the second is negatively charged with q2=-338?C . Use conservation of energy to determine the difference in the speeds of the two balls when they hit the ground. (Neglect air resistance.)
Note that for electric potential energy,
Pq = E q y
And for gravitational potential energy,
Pg = m g y
Thus, as they are dropped, vo = 0, and thus,
1/2mv^2 = mgh + Eqh
Thus,
v = sqrt[2(mgh + Eqh)/m]
Thus, for the first ball,
m = 0.407 kg
q = 3.38E-04 C
h = 2.29 m
Then,
v+ = 6.741997531
m/s
For the negatively charged ball,
q = -3.38E-04 C
Then,
v - = 6.656836282
m/s
Thus, the difference in speeds is
v+ - v - = 0.0852
m/s [ANSWER]