Two charges $+ q_{1}$ and $- q_{2}$ are placed at A and B, respectively. A line of force emanates from $q_{1}$ at an angle $α$ with the line AB. At what angle will it terminate at $- q_{2}$ ?

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$2 \sin^{- 1} [\sin \frac{α}{2} \sqrt{\frac{q_{1}}{q_{2}}}]$

$2 \sin^{- 1} [\sin \frac{α}{2} \sqrt{\frac{q_{2}}{q_{1}}}]$

$2 \cos^{- 1} [\sin \frac{α}{2} \sqrt{\frac{q_{1}}{q_{2}}}]$

$2 \cos^{- 1} [\sin \frac{α}{2} \sqrt{\frac{q_{2}}{q_{1}}}]$

answer is A.

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Detailed Solution

$N_{1} = \frac{q}{4 π} 2 π (1 - \cos α), N_{2} = \frac{q_{2}}{4 π} 2 π (1 - \cos β)$

$\frac{q_{1}}{4 π} 2 π (1 - \cos α) = \frac{q_{2}}{4 π} 2 π (1 - \cos β)$

$β = 2 \sin^{- 1} [\sin \frac{α}{2} \sqrt{\frac{q_{1}}{q_{2}}}]$

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