Answer:
U = 25 J
Explanation:
The energy in a set of charges is given by
U = [tex]k \sum \frac{q_i}{r_i}[/tex]
in this case we have three charges of equal magnitude
q = q₁ = q₂ = q₃
with the configuration of an equilateral triangle all distances are worth
d = a
U = k ( [tex]\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3}[/tex] )
we substitute
15 = k q² (3 / a)
k q² /a = 5
For the second configuration a load is moved to the measured point of the other two
d₁₃ = a
The distance to charge 2 that is at the midpoint of the other two is
d₁₂ = d₂₃ = a / 2
U = k (\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3})
U = k q² ( [tex]\frac{2}{a} + \frac{1}{a} + \frac{2}{a}[/tex] )
U = (kq² /a) 5
substituting
U = 5 5
U = 25 J
