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Three identical charges, each $2 μ C$ lie at the vertices of a right angled triangle as shown in the figure. Forces on the charge at B due to the charges at A and C respectively are $F_{1}$ and $F_{2}$ . The angle between their resultant force and $F_{2}$ is

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$\tan^{- 1} (\frac{9}{7})$

$\tan^{- 1} (\frac{9}{16})$

$\tan^{- 1} (\frac{16}{9})$

$\tan^{- 1} (\frac{7}{9})$

answer is A.

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Detailed Solution

$F_{1} = \frac{1}{4 π \in_{0}} \frac{Q Q}{3^{2}}$

$F_{2} = \frac{1}{4 π \in_{0}} \frac{Q Q}{4^{2}}$

$\tan α = \frac{F_{1}}{F_{2}} \Rightarrow \tan α = \frac{4^{2}}{3^{2}} = \frac{16}{9}$

$∴ α = \tan^{- 1} (\frac{16}{9})$

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