Question

Three identical charges of magnitude $2\mu C$ are placed at the corners of a right angled triangle ABC whose base BC and height BA are respectively $4cm$ and $3cm$. Forces on the charge at the right angled corner B due to the charges at A and C are respectively $F_1$ and $F_2$. The angle between their resultant force and $F_2$ is :

• A. $ta{n}^{-1}\left(\frac{9}{16}\right)$
• B. $ta{n}^{-1}\left(\frac{16}{9}\right)$
• C. $si{n}^{-1}\left(\frac{16}{9}\right)$
• D. $co{s}^{-1}\left(\frac{16}{9}\right)$

Answer 1

The correct option is B $ta{n}^{-1}\left(\frac{16}{9}\right)$

Let the effective force ($F$) angle be at angle $\theta to{F}_2$
We know $Fcos\theta =F_2$ and $Fsin\theta =F_1$
$\Rightarrow tan\theta =\frac{F_1}{F_2}$
$\Rightarrow tan\theta =\frac{\frac{K{q}^2}{(3{)}^2}}{\frac{K{q}^2}{(4{)}^2}}$
$=\frac{4^2}{3^2}$
$tan\theta =\frac{16}{9}$
$\theta =ta{n}^{-1}\left(\frac{16}{9}\right)$