Question

Four identical charges (+2.7 μC each) are brought from infinity and fixed to a straight line. Each charge is 0.42 m from the next. Determine the electric potential energy of this group.

Answer 1

When we bring 1st charge on the line there will be no electric potential energy (EPE) required, since there is no resistance by any other charge, Now then we bring 2nd charge on the line at 0.42 m from 1st charge then

EPE1 = k*q1*q2/r12

Now then we bring 3rd charge on line then energy required will be:

EPE2 = k*q1*q3/r13 + k*q2*q3/r23

Now then we bring 4th charge on line then energy required will be:

EPE3 = k*q1*q4/r14 + k*q2*q4/r24 + k*q3*q4/r34

Now net electric potential energy of the group will be:

EPE_net = EPE1 + EPE2 + EPE3

EPE_net = k*q1*q2/r12 + k*q1*q3/r13 + k*q2*q3/r23 + k*q1*q4/r14 + k*q2*q4/r24 + k*q3*q4/r34

Now given that

q1 = q2 = q3 = q4 = 2.7*10^-6 C

r12 = r23 = r34 = r = 0.42 m

r13 = r24 = 2r = 2*0.42 = 0.84 m

r14 = 3*0.42 = 3r = 1.26 m

Using these values:

EPE_net = (k*q*q/r)*(1 + 1/2 + 1 + 1/3 + 1/2 + 1)

EPE_net = (k*q^2/r)*(13/3)

EPE_net = 13*9*10^9*(2.7*10^-6)^2/(3*0.42)

EPE_net = 0.68 J

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