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Three identical charges of magnitude $2 μC$ are placed at the corners of right angled triangle ABC whose base BC and height BA respectively 4cm and 3cm. 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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$\sin^{- 1} (\frac{3}{4})$

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

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

$45^{0}$

answer is B.

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Detailed Solution

Given Data: Three identical charges are placed at corners of right angled triangle ABC.

Formula Used: $F = k \frac{q_{1} q_{2}}{r^{2}}$

Detailed Solution:

Let the effective force (F) be at angle $θ$ to $F_{2}$

We know that $F \cos θ = F_{2}$ and $F \sin θ = F_{1}$

$\tan θ = \frac{F_{1}}{F_{2}}$

$\tan θ = \frac{\frac{{Kq}^{2}}{(3)^{2}}}{\frac{Kq}{(4)^{2}}} = \frac{4^{2}}{3^{2}}$

$θ = \tan^{- 1} (\frac{16}{9})$

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