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Question
Three charges –q, +Q, and –q are placed at equal distances on a straight line. If the potential energy of the system of the three charges is zero, then what is the ratio of Q: q?
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Solution
In the above figure, The x-axis is illustrated as a line joining the charges, with the origin at the +Q charge. Let q1 = +Q and q2 = q3 = –q. Since the charges are supposed to be equidistant, place the two –q charges at (–a, O) and (a, 0).
∴ r21 = r31 = a and r32 = 2a
The total potential energy of the system of three charges is
u3 = `1/(4piε_0) (("q"_1"q"_2)/"r"_21 + ("q"_1"q"_3)/"r"_31 + ("q"_2"q"_3)/"r"_32)`
`= 1/(4piε_0) [(-"q"("Q"))/"a" + ("Q"(-"q"))/"a" + ((-"q")(-"q"))/"2a"]`
`= 1/(4piε_0) [- (2"qQ")/"a" + "q"^2/"2a"]`
Given: u3 = 0
∴ `(2"qQ")/"a" = "q"^2/"2a"`
∴ `"Q"/"q" = 1/4`
This gives the required ratio.
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