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 0.407kg 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.)

Answer 1

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]