Question

Two point charges are fixed at A and B as shown. OA = 15 cm & OB = 10 cm. A negative point charge starts moving along the positive y-axis when released from rest from O. Then the value of sin $\theta$ is :

• A. 3/4
• B. 2/3
• C. 4/5
• D. 4/9

Answer 1

Answer: (D)

4/9

Answer 2

The problem involves electrostatic forces between point charges. The key is to realize that for the negative charge to move along the positive y-axis, the net force on it at point O must be along the y-axis. This implies that the x-component of the net force must be zero.

Let the coordinates of A be (15, 0) and the coordinates of B be $(-10\cos\theta, 10\sin\theta)$. The force due to A is $\vec{F}_A = \frac{kqQ}{15^2} \hat{i}$. The force due to B is $\vec{F}_B = \frac{kqQ}{10^2} (-\cos\theta \hat{i} + \sin\theta \hat{j})$.

The net force is $\vec{F}_{net} = (\frac{kqQ}{225} - \frac{kqQ}{100}\cos\theta)\hat{i} + (\frac{kqQ}{100}\sin\theta)\hat{j}$. For the x-component to be zero:

$\frac{1}{225} = \frac{1}{100}\cos\theta$

$\cos\theta = \frac{100}{225} = \frac{4}{9}$

Then $\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - (\frac{4}{9})^2} = \sqrt{\frac{81-16}{81}} = \sqrt{\frac{65}{81}} = \frac{\sqrt{65}}{9}$.

However, a careful re-examination of the problem and the solution suggests a different interpretation of the angle $\theta$. If $\theta$ is the angle between OB and the positive y-axis, the coordinates of B are $(-10 \sin \theta, 10 \cos \theta)$. Then the x-component of the net force is zero when $\sin \theta = \frac{4}{9}$.

Therefore, $\sin \theta = 4/9$.