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Q.

The charges –q, Q, –q are placed along x-axis at x=0, x=a and x=2a respectively. If the potential energy of the system is zero, then Q:q is

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a

1 : 4

b

2 : 1

c

1 : 2

d

4 : 1

answer is C.

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Detailed Solution

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U = U₁₂ + U₂₃+U₃₁
$\begin{array}{l} U = \frac{1}{4 {πε}_{0}} [\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{2} q_{3}}{r_{23}} + \frac{q_{3} q_{1}}{r_{31}}] = 0 \\ U = \frac{- qQ}{a} - \frac{qQ}{a} + \frac{q^{2}}{2 a} = 0 \\ \Rightarrow \frac{q^{2}}{2 a} = \frac{2 qQ}{a} \Rightarrow \frac{Q}{q} = 1 : 4 \end{array}$

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