Three identical charges, each $2 \mu \mathrm{C}$ lie at the vertices of a right angled triangle as shown in…

Three identical charges, each $2 \mu \mathrm{C}$ 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

1. $\tan ^{-1}\left(\frac{9}{16}\right)$
2. $\tan ^{-1}\left(\frac{9}{7}\right)$
3. $\tan ^{-1}\left(\frac{16}{9}\right)$
4. $\tan ^{-1}\left(\frac{7}{9}\right)$

Solution

From figure the net force $F_{\text {net }}$ due to $F_1$ and $F_2$ makes an angle $\theta$ with force $\mathrm{F}_2$
$\begin{aligned} & \tan \theta=\frac{\mathrm{F}_1}{\mathrm{~F}_2} \\ & \text { Also, } \mathrm{F}_1=k \cdot \frac{q_1 q_2}{(3)^2} \\ & \because \quad q_1=q_2=q_3=2 \mu \mathrm{C} \\ & \Rightarrow \quad \mathrm{F}_2=k \cdot \frac{q_1 q_3}{(4)^3} \\ & \therefore \quad \tan \theta=k \cdot \frac{q_1 q_2 /(3)^2}{q_1 q_3 /(4)^2}=\frac{(4)^2}{(3)^2}=\frac{16}{9} \\ & \therefore \quad \theta=\tan ^{-1}\left(\frac{16}{9}\right) \end{aligned}$

Asked in: AP EAMCET 2016

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