Three identical charges of magnitude 2 $μ$ C are placed at the corners of right angled triangle ABC whose base BC and height BA respectively 4 cm and 3 cm. Forces on charge at right angled corner B due to charges at ‘A’ and ‘C’ are respectively F₁ and F₂. The angle between their resultant force and F₂ is

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$\cos^{- 1} (\frac{1}{3})$

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

${Sin}^{- 1} (\frac{3}{4})$

45°

answer is B.

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Detailed Solution

$\begin{array}{l} F_{2} = \frac{1}{4 {πε}_{0}} \frac{2 \times 10^{- 6} \times 2 \times 10^{- 6}}{{(4 \times 10^{- 2})}^{2}} \\ = \frac{9 \times 10^{9} \times 4 \times 10^{- 12}}{16 \times 10^{- 4}} = \frac{9}{4} \times 10^{1} \end{array}$
$\begin{array}{l} F_{1} = \frac{1}{4 {πε}_{0}} \frac{2 \times 10^{- 6} \times 2 \times 10^{- 6}}{{(3 \times 10^{- 2})}^{2}} = \frac{9 \times 10^{9} \times 4 \times 10^{- 12}}{9 \times 10^{- 4}} \\ = 4 \times 10^{2}; \tan θ = \frac{F_{1}}{F_{2}} = \frac{4 \times 10^{1}}{9 / 4 \times 10^{1}} = \frac{16}{9} \end{array}$