Four charges are brought from infinity and placed at one meter intervals as shown. Determine the electric potential energy of this group?

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Answer is E, why?

A) 0.068 J
B) 0.032 J
C) 0 J
D) - 0.014 J
E) - 0.021 J

1 Answer
Mar 15, 2018

suppose,the charge placed at origin is #q_1# and next to it we give name as #q_2,q_3,q_4#

Now,potential energy due to two charges of #q_1# and #q_2# separated by distance #x# is #1/(4 pi epsilon) (q_1)(q_2)/x#

So,here potential energy of the system will be,

#9*10^9((q_1 q_2)/1 +(q_1 q_3)/2 +(q_1 q_4)/3 +(q_2 q_3)/1 +(q_2 q_4)/2 +(q_3 q_4)/1)# (i.e sum of potential energy due to all possible charge combination)

#=9*10^9(-1/1 +1/2 +(-1)/3+(-1)/1 +1/2 +(-1)/1)*10^-6*10^-6#

#=9*10^-3*(-7/3)=-0.021J#