Two 1.5g beads, each charged to 5.5nC , are 1.8cm apart. A 3.0g bead charged to -1.0nC is
exactly halfway between them. The beads are released from rest.
A) What is the speed of the positive beads, in , when they are very far apart? (cm/s)
Solution
Initial PE on both the positive charges is same = PE due to Negative charge + PE due to another
positive charge
Therefore initial PE = ( K (5.5 x 10-9)(5.5 x 10-9) / 0.018 ) + ( K (5.5 x 10-9)(-1 x 10-9) / 0.009 )
= 1.748 x 10-6 J
At infinity, PE becomes Zero and it converts to KE
1.748 x 10-6 J = 0.5 x 0.003 x V2
V = 0.0341 m/sec
exactly halfway between them. The beads are released from rest.
A) What is the speed of the positive beads, in , when they are very far apart? (cm/s)
Solution
Initial PE on both the positive charges is same = PE due to Negative charge + PE due to another
positive charge
Therefore initial PE = ( K (5.5 x 10-9)(5.5 x 10-9) / 0.018 ) + ( K (5.5 x 10-9)(-1 x 10-9) / 0.009 )
= 1.748 x 10-6 J
At infinity, PE becomes Zero and it converts to KE
1.748 x 10-6 J = 0.5 x 0.003 x V2
V = 0.0341 m/sec
