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, $F_{1}=k \cdot \frac{q_{1}\, q_{2}}{(3)^{2}}$
$\because q_{1}=q_{2}=q_{3}=2 \mu C$
$\Rightarrow F_{2}=k \cdot \frac{q_{1}\, q_{3}}{(4)^{2}}$
$\therefore \tan \theta=\frac{k \cdot q_{1} \,q_{2} /(3)^{2}}{k \cdot q_{1}\, q_{3} /(4)^{2}}=\frac{(4)^{2}}{(3)^{2}}=\frac{16}{9}$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{16}{9}\right)$