Question

In: Physics

Near the surface of the Earth there is an electric field of about 150 V/m which...

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.)

Solutions

Expert Solution

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]


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