Question

A charged ball of mass $900\text{g}$ is suspended from a string in a uniform electric field $\overrightarrow{E}=(3\hat{i}+5\hat{j})\times {10}^5\text{N/C}$. The ball is in equilibrium with $\theta ={37}^{\circ }$. If the direction of electric field is reversed, find the new equilibrium position of the ball. Give your answer in terms of angle made by string with vertical. Take $g=10{\text{m/s}}^2$

• A. ${\tan }^{-1}\left(\frac{3}{7}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{14}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{3}\right)$
• D. ${\tan }^{-1}\left(\frac{14}{3}\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(\frac{3}{14}\right)$

Given:
$m=900\text{g}=0.9\text{kg};\theta ={37}^{\circ }$
$\overrightarrow{E}=(3\hat{i}+5\hat{j})\times {10}^5\text{N/C}$

From the diagram given in the question, we can see that the direction of displacement of charge is along the electric field, so charge must be positive in nature.

Drawing an F.B.D of charged ball before reversal of electric field is as shown below,


From the diagram,
$T\sin \theta =3q\times {10}^5$ and
$T\cos \theta +(5q\times {10}^5)=mg$

Solving the above equations we get,

$q=10\mu \text{C}$ and $T=5\text{N}$

After the reversal of direction of electric field. Let $T^{\prime }$ be the tension in the string and $\alpha$ be the angle made by string with vertical.



We get,
$T^{\prime }\sin \alpha =3q\times {10}^5$ and $T^{\prime }\cos \alpha =mg+5q\times {10}^5$

Thus, $\tan \alpha =\frac{3q\times {10}^5}{mg+5q\times {10}^5}$

$\tan \alpha =\frac{3\times {10}^{-5}\times {10}^5}{(0.9\times 10)+(5\times {10}^{-5}\times {10}^5)}=\frac{3}{14}$

$\therefore \alpha ={\tan }^{-1}\left(\frac{3}{14}\right)$

Hence, option $\left(\text{B}\right)$ is correct.