A charge of #2# #C# is at the origin. How much energy would be applied to or released from a #-3# #C# charge if it is moved from # ( 6 , 4 ) # to #(-3 ,2 ) #?

1 Answer
Aug 5, 2016

Answer:

The key to this problem is to find the change in the distance between the two charges. The change in energy will be given by #U=(kq_1q_2)/r#. The energy lost to the environment is #1.5xx10^10# #J#.

Explanation:

Initial separation:

#r_1=sqrt((6-0)^2+(4-0)^2)=sqrt52~~7.2# #m# (we have to assume the units are #m#)

Final separation:

#r_2=sqrt((-3-0)^2+(2-0)^2)=sqrt13~~3.6# #m#

Difference: #7.2-3.6 = 3.6# #m#

The change in the energy will be:

#DeltaU = (kq_1q_2)/(Delta r) = (9xx10^9xx2xx-3)/3.6 = -1.5xx10^10# #J#.

Since the charges are opposite and they have become closer, the energy of the charge system has decreased, so energy will have been lost to the environment.