a. PE=kqq/r
b. the speed will depend on the KE.
KE=OriginalPE-finalPE
=kqq(1/.9 -1/.66)
(q=(9micro, -.5micro)
then 1/2 mv^2=above, solve for v.
a point charge q1=+9μc is held fixed at the origin. A second point charge with a charge of-5μC and mass of 3.2 x 10^-4 kg is placed on the + x-axis, 0.90 m from the origin. a.) What is the potential energy of the pair of charges? (consider the potential to be zero at infinite.) b.) if the second point charge is released from rest, what is it's speed 0.24 m from the origin along the x-axis?
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1 answer