Question

Three identical charges, each $2\mu C$ 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 belween their resultanl force and $F_2$ is

• A. ${\tan }^{-1}\left(\frac{9}{16}\right)$
• B. ${\tan }^{-1}\left(\frac{9}{7}\right)$
• C. ${\tan }^{-1}\left(\frac{16}{9}\right)$
• D. ${\tan }^{-1}\left(\frac{7}{9}\right)$

Answer 1

Answer: (C)

According to the question, forces $F_1$ and $F_2$ are acting at point $B$.

Let the net force $F_{\text{net }}$ due to $F_1$ and $F_2$ makes an angle $\theta$ with force $F_2$.
From figure, $\tan \theta =\frac{F_1}{F_2}$
Also, $F_1=k\cdot \frac{q_1{q}_2}{(3{)}^2}$
$\because q_1=q_2=q_3=2\mu C$
$\Rightarrow F_2=k\cdot \frac{q_1{q}_3}{(4{)}^2}$
$\therefore \tan \theta =\frac{k\cdot q_1{q}_2/(3{)}^2}{k\cdot q_1{q}_3/(4{)}^2}=\frac{(4{)}^2}{(3{)}^2}=\frac{16}{9}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{16}{9}\right)$