Question

Three particles, charge q₁ = +12 µC, q₂ = -19 µC, and q₃ = +31 µC, are positioned at the vertices of an isosceles triangle as shown in the figure. If a = 10 cm and b = 5.7 cm, how much work must an external agent do to exchange the positions of (a)  q₁ and q₃ and, instead, (b)  q₁ and q₂?

Answer 1

q1 = +12 µC,
q2 = -19 µC
q3 = +31 µC
a = 10 cm = 0.10 m
b = 5.7 cm = 0.057 m

Potential Energy of a system is given by = k ∑ (q1q2/b + q1q3/a + q2q3/a)

Initial Potential Energy = 8.9*10^-9 * ( -(12*19)/0.057 + (12*31)/0.10 - (19*31)/0.10)
Initial Potential Energy = -5.5 * 10^5 J

(a)
When q1 and q3 exchange the positions -
Potential Energy = 8.9*10^-9 * ( -12*19/0.10- 12*31/0.10 - 19*31/0.057)
P.E final = -7.9 * 10^5 J

Work done by External Agent = Change in Potential Energy.
Work done by External Agent = Final Potential Energy - Initial Potential Energy
Work done by External Agent = -7.9 * 10^5 J + -5.5 * 10^5 J
Work done by External Agent = -2.4 * 10^5 J

(b)
When q1 and q2 exchange the positions -
Potential Energy = 8.9*10^-9 * ( -(12*19)/0.057 + (12*31)/0.10 - (19*31)/0.10)
P.E final =  -5.5 * 10^5 J

Work done by External Agent = Change in Potential Energy.
Work done by External Agent = Final Potential Energy - Initial Potential Energy
Work done by External Agent =  -5.5 * 10^5 J -  -5.5 * 10^5 J
Work done by External Agent =  0