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Chinese Revolution

Three point c...

Question

Three point charges q, -2q and -2q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to 2a will be

\frac{1}{4 π ϵ_{0}} . \frac{2 q^{2}}{a}

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\frac{1}{4 π ϵ_{0}} . \frac{3 q^{2}}{a}

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Solution

The correct option is C $Z e r o$
$U_{i} = \frac{1}{4 π ϵ_{0}} . [\frac{q_{1} q_{2}}{a} + \frac{q_{2} q_{3}}{a} + \frac{q_{3} q_{1}}{a}] = \frac{1}{4 π ϵ_{0}} [\frac{- 2 q_{2}}{a} + \frac{4 q^{2}}{a} - \frac{2 q^{2}}{a}] = 0 U_{f} = 0 W = U_{f} - U_{i} = 0$

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Three point charges q, -2q and -2q are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to 2a will be

The correct option is C ZeroUi=14πϵ0.[q1q2a+q2q3a+q3q1a]=14πϵ0[−2q2a+4q2a−2q2a]=0Uf=0W=Uf−Ui=0

The correct option is C $Z e r o$
$U_{i} = \frac{1}{4 π ϵ_{0}} . [\frac{q_{1} q_{2}}{a} + \frac{q_{2} q_{3}}{a} + \frac{q_{3} q_{1}}{a}] = \frac{1}{4 π ϵ_{0}} [\frac{- 2 q_{2}}{a} + \frac{4 q^{2}}{a} - \frac{2 q^{2}}{a}] = 0 U_{f} = 0 W = U_{f} - U_{i} = 0$